Luy thua (tu Han-Viet: 累乘  nghia la "nhan chong chat len") la mot phep toan toan hoc, đuoc viet duoi dang an, bao gom hai so, co so a va so mu hoac luy thua n, va đuoc phat am la "a luy thua n". Khi n la mot so nguyen duong, luy thua tuong ung voi phep nhan lap cua co so (thua so): nghia la an la tich cua phep nhan n co so:

  
    
      
        
          a
          
            n
          
        
        =
        
          
            
              
                a
                ×
                ⋯
                ×
                a
              
              ⏟
            
          
          
            n
            
            
              

              
            
          
        
        .
      
    
    {\displaystyle a^{n}=\underbrace {a\times \dots \times a} _{n\,{\textrm {}}}.}
  

So mu thuong đuoc hien thi duoi dang chi so tren o ben phai cua co so. Trong truong hop đo.

an đuoc goi la "luy thua bac n cua a", "a luy thua n", hoac hau het ngan gon la "a mu n"

  
    
      
        
          a
          
            2
          
        
      
    
    {\displaystyle a^{2}}
  
 con đuoc goi la "a binh phuong" hoac "binh phuong cua a"

  
    
      
        
          a
          
            3
          
        
      
    
    {\displaystyle a^{3}}
  
 con đuoc goi la "a lap phuong" hoac "lap phuong cua a"
Ta co a1 = a, va, voi moi so nguyen duong m va n, ta co am ⋅ an = am+n. Đe mo rong thuoc tinh nay thanh so mu nguyen khong duong, a0 (voi a khac 0) đuoc đinh nghia la 1, a−n (voi n la so nguyen duong va a khac 0) đuoc đinh nghia la 1/an. Đac biet, a−1 bang 1/a, nghich đao cua a.
Đinh nghia ve luy thua co the đuoc mo rong đe cho phep bat ky so mu thuc hoac phuc nao. Luy thua theo so mu nguyen cung co the đuoc đinh nghia cho nhieu loai cau truc đai so, bao gom ca ma tran.
Luy thua đuoc su dung rong rai trong nhieu linh vuc, bao gom kinh te hoc, sinh hoc, hoa hoc, vat ly va khoa hoc may tinh, voi cac ung dung nhu lai kep, tang dan so, đong hoc phan ung hoa hoc, hanh vi song va mat ma khoa cong khai.


Luy thua voi so mu nguyen = Luy thua cua 0 va 1 =

  
    
      
        
          0
          
            n
          
        
        =
        0
        
      
    
    {\displaystyle 0^{n}=0\,}
  

  
    
      
        (
        n
        >
        0
        )
        
      
    
    {\displaystyle (n>0)\,}
  

  
    
      
        
          1
          
            n
          
        
        =
        1
        
      
    
    {\displaystyle 1^{n}=1\,}
  


= Luy thua voi so mu nguyen duong =
Luy thua bac n cua a la tich cua n thua so bang nhau, moi thua so bang a:

  
    
      
        
          a
          
            n
          
        
        =
        
          
            
              
                a
                ×
                a
                ×
                ⋯
                ×
                a
              
              ⏟
            
          
          
            n
          
        
      
    
    {\displaystyle a^{n}=\underbrace {a\times a\times \cdots \times a} _{n}}
  

Cac tinh chat quan trong nhat cua luy thua voi so mu nguyen duong m, n la

  
    
      
        
          a
          
            m
            +
            n
          
        
        =
        
          a
          
            m
          
        
        ×
        
          a
          
            n
          
        
      
    
    {\displaystyle a^{m+n}=a^{m}\times a^{n}}
  

  
    
      
        
          a
          
            m
            −
            n
          
        
        =
        
          
            
              a
              
                m
              
            
            
              a
              
                n
              
            
          
        
      
    
    {\displaystyle a^{m-n}={\frac {a^{m}}{a^{n}}}}
  
 
  
    
      
        (
        ∀
        a
        =
        0
        )
      
    
    {\displaystyle (\forall a\neq 0)}
  

  
    
      
        (
        
          a
          
            m
          
        
        
          )
          
            n
          
        
        =
        
          a
          
            m
            n
          
        
      
    
    {\displaystyle (a^{m})^{n}=a^{mn}}
  

  
    
      
        
          a
          
            
              m
              
                n
              
            
          
        
        =
        
          a
          
            (
            
              m
              
                n
              
            
            )
          
        
      
    
    {\displaystyle a^{m^{n}}=a^{(m^{n})}}
  

  
    
      
        (
        a
        b
        
          )
          
            n
          
        
        =
        
          a
          
            n
          
        
        ⋅
        
          b
          
            n
          
        
      
    
    {\displaystyle (ab)^{n}=a^{n}\cdot b^{n}}
  

  
    
      
        
          
            (
            
              
                a
                b
              
            
            )
          
          
            n
          
        
        =
        
          
            
              a
              
                n
              
            
            
              b
              
                n
              
            
          
        
      
    
    {\displaystyle \left({\frac {a}{b}}\right)^{n}={\frac {a^{n}}{b^{n}}}}
  

Đac biet, ta co:

  
    
      
        
          a
          
            1
          
        
        =
        a
      
    
    {\displaystyle a^{1}=a}
  

Trong khi cac phep cong va phep nhan co tinh chat giao hoan, phep tinh luy thua khong co tinh giao hoan.
Tuong tu cac phep cong va nhan co tinh ket hop, con phep tinh luy thua thi khong. Khi khong co dau ngoac, thu tu tinh cua cac luy thua la tu tren xuong, chu khong phai la tu duoi len:

  
    
      
        
          a
          
            
              m
              
                n
              
            
          
        
        =
        
          a
          
            (
            
              m
              
                n
              
            
            )
          
        
        =
        (
        
          a
          
            m
          
        
        
          )
          
            n
          
        
        =
        
          a
          
            (
            m
            n
            )
          
        
        =
        
          a
          
            m
            n
          
        
      
    
    {\displaystyle a^{m^{n}}=a^{(m^{n})}\neq (a^{m})^{n}=a^{(mn)}=a^{mn}}
  

Luy thua bac chan cua mot so am la so duong.
Luy thua bac le cua mot so am la so am.


= Luy thua voi so mu 0 =
Luy thua voi so mu 0 cua so a = 0 đuoc quy uoc bang 1. 

  
    
      
        
          a
          
            0
          
        
        =
        1
      
    
    {\displaystyle a^{0}=1}
  

Chung minh:

  
    
      
        1
        =
        
          
            
              a
              
                n
              
            
            
              a
              
                n
              
            
          
        
        =
        
          a
          
            n
            −
            n
          
        
        =
        
          a
          
            0
          
        
      
    
    {\displaystyle 1={\frac {a^{n}}{a^{n}}}=a^{n-n}=a^{0}}
  


= Luy thua voi so mu nguyen am =
Luy thua cua a voi so mu nguyen am -n, a khac 0 va n la so nguyen duong la: 

  
    
      
        
          a
          
            −
            n
          
        
        =
        
          
            1
            
              a
              
                n
              
            
          
        
      
    
    {\displaystyle a^{-n}={\frac {1}{a^{n}}}}
  
.
Vi du 

  
    
      
        
          3
          
            −
            4
          
        
        =
        
          
            1
            
              3
              
                4
              
            
          
        
        =
        
          
            1
            
              3
              ⋅
              3
              ⋅
              3
              ⋅
              3
            
          
        
        =
        
          
            1
            81
          
        
      
    
    {\displaystyle 3^{-4}={\frac {1}{3^{4}}}={\frac {1}{3\cdot 3\cdot 3\cdot 3}}={\frac {1}{81}}}
  
.
Cach suy luan ra "luy thua voi so mu nguyen am" tu "luy thua voi so mu 0":

  
    
      
        
          a
          
            0
          
        
        =
        
          a
          
            n
            −
            n
          
        
        =
        
          
            
              a
              
                n
              
            
            
              a
              
                n
              
            
          
        
        =
        
          a
          
            n
          
        
        ⋅
        
          
            1
            
              a
              
                n
              
            
          
        
        =
        
          a
          
            n
          
        
        ⋅
        
          a
          
            −
            n
          
        
      
    
    {\displaystyle a^{0}=a^{n-n}={\frac {a^{n}}{a^{n}}}=a^{n}\cdot {\frac {1}{a^{n}}}=a^{n}\cdot a^{-n}}
  

Truong hop đac biet: luy thua cua so a = 0 voi so mu −1 la so nghich đao cua no.

  
    
      
        
          a
          
            −
            1
          
        
        =
        
          
            1
            a
          
        
        .
      
    
    {\displaystyle a^{-1}={\frac {1}{a}}.}
  


Luy thua cua so thuc duong voi so mu huu ty = Can bac n cua mot so thuc duong =
Mot can bac n cua so a la mot so x sao cho xn = a.
Neu a la so thuc duong, n la so nguyen duong thi co đung mot so thuc duong x sao cho xn = a.
So x nay đuoc goi la can so hoc bac n cua a. No đuoc ky hieu la n√a, trong đo  √  la ky hieu can.


= Luy thua voi so mu huu ty cua so thuc duong =
Luy thua voi so mu huu ty toi gian b/c (b, c la so nguyen, trong đo c duong), cua so thuc duong a đuoc đinh nghia la

  
    
      
        
          a
          
            
              b
              c
            
          
        
        =
        (
        
          a
          
            b
          
        
        
          )
          
            
              1
              c
            
          
        
        =
        
          
            
              a
              
                b
              
            
            
              c
            
          
        
      
    
    {\displaystyle a^{\frac {b}{c}}=(a^{b})^{\frac {1}{c}}={\sqrt[{c}]{a^{b}}}}
  

Luy thua voi so mu huu ti, voi co so am la khong co nghia.


Luy thua voi so mu thuc = Luy thua cua so e =
So e la hang so toan hoc quan trong, xap xi 2.718 va la co so cua logarit tu nhien. So e đuoc đinh nghia qua gioi han sau: 

  
    
      
        e
        =
        
          lim
          
            n
            →
            ∞
          
        
        
          
            (
            
              1
              +
              
                
                  1
                  n
                
              
            
            )
          
          
            n
          
        
        .
      
    
    {\displaystyle e=\lim _{n\rightarrow \infty }\left(1+{\frac {1}{n}}\right)^{n}.}
  

Ham e mu, đuoc đinh nghia boi 

  
    
      
        
          e
          
            x
          
        
        =
        
          lim
          
            n
            →
            ∞
          
        
        
          
            (
            
              1
              +
              
                
                  x
                  n
                
              
            
            )
          
          
            n
          
        
        ,
      
    
    {\displaystyle e^{x}=\lim _{n\rightarrow \infty }\left(1+{\frac {x}{n}}\right)^{n},}
  

o đay x đuoc viet nhu so mu vi no thoa man đang thuc co ban cua luy thua

  
    
      
        
          e
          
            x
            +
            y
          
        
        =
        
          e
          
            x
          
        
        ⋅
        
          e
          
            y
          
        
        .
      
    
    {\displaystyle e^{x+y}=e^{x}\cdot e^{y}.}
  

Ham e mu xac đinh voi tat ca cac gia tri nguyen, huu ty, thuc va ca gia tri phuc cua x.
Co the chung minh ngan gon rang ham e mu voi x la so nguyen duong k chinh la ek nhu sau:

  
    
      
        (
        e
        
          )
          
            k
          
        
        =
        
          
            [
            
              
                lim
                
                  n
                  →
                  ∞
                
              
              
                
                  (
                  
                    1
                    +
                    
                      
                        1
                        n
                      
                    
                  
                  )
                
                
                  n
                
              
            
            ]
          
          
            k
          
        
        =
        
          lim
          
            n
            →
            ∞
          
        
        
          
            [
            
              
                (
                
                  1
                  +
                  
                    
                      1
                      n
                    
                  
                
                )
              
              
                n
              
            
            ]
          
          
            k
          
        
        =
        
          lim
          
            n
            →
            ∞
          
        
        
          
            (
            
              1
              +
              
                
                  k
                  
                    n
                    ⋅
                    k
                  
                
              
            
            )
          
          
            n
            ⋅
            k
          
        
      
    
    {\displaystyle (e)^{k}=\left[\lim _{n\rightarrow \infty }\left(1+{\frac {1}{n}}\right)^{n}\right]^{k}=\lim _{n\rightarrow \infty }\left[\left(1+{\frac {1}{n}}\right)^{n}\right]^{k}=\lim _{n\rightarrow \infty }\left(1+{\frac {k}{n\cdot k}}\right)^{n\cdot k}}
  

  
    
      
        =
        
          lim
          
            n
            ⋅
            k
            →
            ∞
          
        
        
          
            (
            
              1
              +
              
                
                  k
                  
                    n
                    ⋅
                    k
                  
                
              
            
            )
          
          
            n
            ⋅
            k
          
        
        =
        
          lim
          
            m
            →
            ∞
          
        
        
          
            (
            
              1
              +
              
                
                  k
                  m
                
              
            
            )
          
          
            m
          
        
        =
        
          e
          
            k
          
        
        .
      
    
    {\displaystyle =\lim _{n\cdot k\rightarrow \infty }\left(1+{\frac {k}{n\cdot k}}\right)^{n\cdot k}=\lim _{m\rightarrow \infty }\left(1+{\frac {k}{m}}\right)^{m}=e^{k}.}
  

Chung minh nay cung chung to rang ex+y thoa man đang thuc luy thua khi x va y la cac so nguyen duong. Ket qua nay cung co the mo rong cho tat ca cac so khong phai la so nguyen duong.


= Luy thua voi so mu thuc =
Vi moi so thuc co the đuoc tiem can boi cac so huu ty nen luy thua cua voi so mu thuc x co the đinh nghia nho gioi han

  
    
      
        
          b
          
            x
          
        
        =
        
          lim
          
            r
            →
            x
          
        
        
          b
          
            r
          
        
        ,
      
    
    {\displaystyle b^{x}=\lim _{r\to x}b^{r},}
  

trong đo r tien toi x chi tren cac gia tri huu ty cua r.
Chang han, neu 

  
    
      
        x
        ≈
        1
        ,
        732
      
    
    {\displaystyle x\approx 1,732}
  

thi 

  
    
      
        
          5
          
            x
          
        
        ≈
        
          5
          
            1
            ,
            732
          
        
        =
        
          5
          
            433
            
              /
            
            250
          
        
        =
        
          
            
              5
              
                433
              
            
            
              250
            
          
        
        ≈
        16
        ,
        241.
      
    
    {\displaystyle 5^{x}\approx 5^{1,732}=5^{433/250}={\sqrt[{250}]{5^{433}}}\approx 16,241.}
  

Luy thua voi so mu thuc cung thuong đuoc đinh nghia bang cach su dung logarit thay cho su dung gioi han cua cac so huu ty.
Logarit tu nhien 
  
    
      
        ln
        ⁡
        
          (
          x
          )
        
      
    
    {\displaystyle \ln {(x)}}
  
 la ham nguoc cua ham e-mu ex. Theo đo 
  
    
      
        ln
        ⁡
        x
      
    
    {\displaystyle \ln x}
  
 la so b sao cho x = eb .
Neu a la so thuc duong, x la so thuc bat ky ta co a = e  ln a
nen neu ax đuoc đinh nghia nho ham logarit tu nhien thi ta can phai co

  
    
      
        
          a
          
            x
          
        
        =
        (
        
          e
          
            ln
            ⁡
            a
          
        
        
          )
          
            x
          
        
        =
        
          e
          
            x
            ⋅
            ln
            ⁡
            a
          
        
        .
        
      
    
    {\displaystyle a^{x}=(e^{\ln a})^{x}=e^{x\cdot \ln a}.\,}
  

Đieu nay dan toi đinh nghia

  
    
      
        
          a
          
            x
          
        
        =
        
          e
          
            x
            ⋅
            ln
            ⁡
            a
          
        
        
      
    
    {\displaystyle a^{x}=e^{x\cdot \ln a}\,}
  

voi moi so thuc x va so thuc duong a.
Đinh nghia nay cua luy thua so mu thuc phu hop voi đinh nghia luy thua thuc nho gioi han o tren va voi ca luy thua voi so mu phuc duoi đay.


Luy thua voi so mu phuc = Luy thua so mu phuc cua so e =
Dua vao bieu dien luong giac cua cac so phuc, nguoi ta đinh nghia luy thua so mu phuc cua so e nhu sau.
Truoc het, luy thua voi so mu thuan ao cua e đinh nghia theo cong thuc Euler:

  
    
      
        
          e
          
            i
            x
          
        
        =
        cos
        ⁡
        x
        +
        i
        ⋅
        sin
        ⁡
        x
      
    
    {\displaystyle e^{ix}=\cos x+i\cdot \sin x}
  

Sau đo voi so phuc 
  
    
      
        z
        =
        x
        +
        y
        ⋅
        i
      
    
    {\displaystyle z=x+y\cdot i}
  
, ta co

  
    
      
        
          e
          
            z
          
        
        =
        
          e
          
            x
            +
            i
            y
          
        
        =
        
          e
          
            x
          
        
        ⋅
        
          e
          
            i
            y
          
        
        =
        
          e
          
            x
          
        
        (
        cos
        ⁡
        y
        +
        i
        ⋅
        sin
        ⁡
        y
        )
      
    
    {\displaystyle e^{z}=e^{x+iy}=e^{x}\cdot e^{iy}=e^{x}(\cos y+i\cdot \sin y)}
  


= Luy thua so mu phuc cua so thuc duong =
Neu a la mot so thuc duong va z la so phuc thi luy thua az đuoc đinh nghia la

  
    
      
        
          a
          
            z
          
        
        =
        
          
            
              
                (
              
            
            
              e
              
                ln
                ⁡
                a
              
            
            
              
                )
              
            
          
          
            z
          
        
        =
        
          e
          
            z
            ⋅
            ln
            ⁡
            a
          
        
      
    
    {\displaystyle a^{z}={{\big (}e^{\ln a}{\big )}}^{z}=e^{z\cdot \ln a}}
  

trong đo x = ln(a) la nghiem duy nhat cua phuong trinh ex = a.
Neu 
  
    
      
        z
        =
        x
        +
        y
        ⋅
        i
      
    
    {\displaystyle z=x+y\cdot i}
  
, ta co

  
    
      
        
          a
          
            z
          
        
        =
        
          e
          
            ln
            ⁡
            a
            ⋅
            (
            x
            +
            i
            y
            )
          
        
        =
      
    
    {\displaystyle a^{z}=e^{\ln a\cdot (x+iy)}=}
  
 
  
    
      
        
          e
          
            x
            ln
            ⁡
            a
            +
            i
            ⋅
            y
            ln
            ⁡
            a
          
        
      
    
    {\displaystyle e^{x\ln a+i\cdot y\ln a}}
  

  
    
      
        =
        
          e
          
            x
            ⋅
            ln
            ⁡
            a
          
        
        ⋅
        
          
            [
          
        
        cos
        ⁡
        (
        y
        ln
        ⁡
        a
        )
        +
        i
        ⋅
        sin
        ⁡
        (
        y
        ln
        ⁡
        a
        )
        
          
            ]
          
        
      
    
    {\displaystyle =e^{x\cdot \ln a}\cdot {\big [}\cos(y\ln a)+i\cdot \sin(y\ln a){\big ]}}
  

  
    
      
        =
        
          a
          
            x
          
        
        ⋅
        
          
            [
          
        
        cos
        ⁡
        (
        y
        ln
        ⁡
        a
        )
        +
        i
        ⋅
        sin
        ⁡
        (
        y
        ln
        ⁡
        a
        )
        
          
            ]
          
        
      
    
    {\displaystyle =a^{x}\cdot {\big [}\cos(y\ln a)+i\cdot \sin(y\ln a){\big ]}}
  


Tinh chat luy thua = Tinh chat co ban =
1) 
  
    
      
        
          a
          
            n
          
        
        =
        a
        ×
        a
        ×
        ⋯
        ×
        a
      
    
    {\displaystyle a^{n}=a\times a\times \dots \times a}
  
 (n thua so a)
2) 
  
    
      
        
          a
          
            −
            n
          
        
        =
        
          
            1
            
              a
              
                n
              
            
          
        
        =
        
          
            1
            
              a
              ×
              a
              ×
              a
              ×
              .
              .
              .
              ×
              a
            
          
        
      
    
    {\displaystyle a^{-n}={\frac {1}{a^{n}}}={\frac {1}{a\times a\times a\times ...\times a}}}
  

3) 
  
    
      
        
          0
          
            n
          
        
        =
        0
        
        (
        n
        =
        0
        )
      
    
    {\displaystyle 0^{n}=0\,(n\neq 0)}
  

4) 
  
    
      
        
          1
          
            n
          
        
        =
        1
      
    
    {\displaystyle 1^{n}=1}
  

5) 
  
    
      
        
          a
          
            0
          
        
        =
        1
        
        (
        a
        =
        0
        )
      
    
    {\displaystyle a^{0}=1\,(a\neq 0)}
  
 
6) 
  
    
      
        
          a
          
            1
          
        
        =
        a
      
    
    {\displaystyle a^{1}=a}
  

7) 
  
    
      
        
          a
          
            −
            1
          
        
        =
        
          
            1
            a
          
        
      
    
    {\displaystyle a^{-1}={\frac {1}{a}}}
  


= Tinh chat thuong gap =
1) 
  
    
      
        
          a
          
            m
            +
            n
          
        
        =
        
          
            a
            
              m
            
          
        
        ×
        
          
            a
            
              n
            
          
        
      
    
    {\displaystyle a^{m+n}={a^{m}}\times {a^{n}}}
  

2) 
  
    
      
        
          a
          
            m
            −
            n
          
        
        =
        
          
            
              a
              
                m
              
            
            
              a
              
                n
              
            
          
        
      
    
    {\displaystyle a^{m-n}={\frac {a^{m}}{a^{n}}}}
  
 
  
    
      
        (
        ∀
        a
        =
        0
        )
      
    
    {\displaystyle (\forall a\neq 0)}
  

3) 
  
    
      
        
          a
          
            m
            ⋅
            n
          
        
        =
        (
        
          a
          
            m
          
        
        
          )
          
            n
          
        
      
    
    {\displaystyle a^{m\cdot n}=(a^{m})^{n}}
  

4) 
  
    
      
        
          a
          
            
              m
              
                n
              
            
          
        
        =
        
          a
          
            (
            
              m
              
                n
              
            
            )
          
        
      
    
    {\displaystyle a^{m^{n}}=a^{(m^{n})}}
  

5) 
  
    
      
        (
        a
        ×
        b
        
          )
          
            n
          
        
        =
        
          a
          
            n
          
        
        ×
        
          b
          
            n
          
        
      
    
    {\displaystyle (a\times b)^{n}=a^{n}\times b^{n}}
  

6)
  
    
      
        
          
            (
            
              
                a
                b
              
            
            )
          
          
            n
          
        
        =
        
          
            
              a
              
                n
              
            
            
              b
              
                n
              
            
          
        
      
    
    {\displaystyle \left({\frac {a}{b}}\right)^{n}={\frac {a^{n}}{b^{n}}}}
  

7) 
  
    
      
        
          a
          
            
              b
              c
            
          
        
        =
        
          
            (
            
              a
              
                b
              
            
            )
          
          
            1
            
              /
            
            c
          
        
        =
        
          
            
              a
              
                b
              
            
            
              c
            
          
        
      
    
    {\displaystyle a^{\frac {b}{c}}=\left(a^{b}\right)^{1/c}={\sqrt[{c}]{a^{b}}}}
  

8) 
  
    
      
        
          a
          
            x
          
        
        =
        
          e
          
            x
            ⋅
            ln
            ⁡
            a
          
        
        
      
    
    {\displaystyle a^{x}=e^{x\cdot \ln a}\,}
  

9) 
  
    
      
        
          e
          
            i
            x
          
        
        =
        cos
        ⁡
        x
        +
        i
        ⋅
        sin
        ⁡
        x
      
    
    {\displaystyle e^{ix}=\cos x+i\cdot \sin x}
  


Ham so luy thua Ham so luy thua la ham so co dang 
  
    
      
        y
        =
        
          x
          
            α
          
        
      
    
    {\displaystyle y=x^{\alpha }}
  
 voi 
  
    
      
        α
        ∈
        
          R
        
      
    
    {\displaystyle \alpha \in \mathbb {R} }
  


= Tap xac đinh =
Tap xac đinh cua ham so tren phu thuoc vao so mu 
  
    
      
        α
      
    
    {\displaystyle \alpha }
  

neu 
  
    
      
        α
      
    
    {\displaystyle \alpha }
  
 la so nguyen duong thi tap xac đinh la 
  
    
      
        D
        =
        
          R
        
      
    
    {\displaystyle D=\mathbb {R} }
  

neu 
  
    
      
        α
        =
        0
      
    
    {\displaystyle \alpha =0}
  
 hoac 
  
    
      
        α
      
    
    {\displaystyle \alpha }
  
 la so nguyen am thi tap xac đinh la 
  
    
      
        D
        =
        
          R
        
        ∖
        {
        0
        }
      
    
    {\displaystyle D=\mathbb {R} \setminus \{0\}}
  

neu 
  
    
      
        α
      
    
    {\displaystyle \alpha }
  
 khong phai la so nguyen thi tap xac đinh la 
  
    
      
        D
        =
        (
        0
        ;
        +
        ∞
        )
      
    
    {\displaystyle D=(0;+\infty )}
  


= Đao ham =
Ham so 
  
    
      
        y
        =
        f
        (
        x
        )
        =
        
          x
          
            α
          
        
      
    
    {\displaystyle y=f(x)=x^{\alpha }}
  
co đao ham tai moi x > 0 va 
  
    
      
        
          y
          ′
        
        =
        α
        
          x
          
            α
            −
            1
          
        
      
    
    {\displaystyle y'=\alpha x^{\alpha -1}}
  
 la đao ham cap 1 cua f(x)


= Chieu bien thien cua ham so luy thua voi bien so duong =
Xet ham so 
  
    
      
        y
        =
        
          x
          
            α
          
        
      
    
    {\displaystyle y=x^{\alpha }}
  
 tren x>0:

Voi 
  
    
      
        α
        >
        0
      
    
    {\displaystyle \alpha >0}
  
, ham so đong bien tren 
  
    
      
        (
        0
        ;
        +
        ∞
        )
      
    
    {\displaystyle (0;+\infty )}
  

Voi 
  
    
      
        α
        <
        0
      
    
    {\displaystyle \alpha <0}
  
, ham so nghich bien tren 
  
    
      
        (
        0
        ;
        +
        ∞
        )
      
    
    {\displaystyle (0;+\infty )}
  


= Đo thi =


 Đo thi ham so luy thua voi so mu thuc va bien so duong  Đo thi ham so 
  
    
      
        y
        =
        
          x
          
            α
          
        
      
    
    {\displaystyle y=x^{\alpha }}
  
tren x>0 co tinh chat sau:

Luon đi qua điem I(1;1)
Neu 
  
    
      
        α
        <
        0
      
    
    {\displaystyle \alpha <0}
  
, đo thi nhan truc Ox la tiem can ngang va truc Oy la tiem can đung
Co đuong bieu dien phu thuoc vao so mu 
  
    
      
        α
      
    
    {\displaystyle \alpha }
  


 Đo thi ham so luy thua voi so mu nguyen  Đo thi ham so 
  
    
      
        y
        =
        f
        (
        x
        )
        =
        
          x
          
            n
          
        
      
    
    {\displaystyle y=f(x)=x^{n}}
  
 voi 
  
    
      
        n
        ∈
        
          Z
        
      
    
    {\displaystyle n\in \mathbb {Z} }
  
 co tinh chat tuong tu nhu tren voi x>0. Ngoai ra, phan đo thi voi x<0 co tinh đoi xung voi phan đo thi x>0 phu thuoc vao n:

Neu n la so chan, đo thi đoi xung qua truc Oy do f(x) la ham so chan
Neu n la so le, đo thi đoi xung qua goc toa đo O do f(x) la ham so le


Ham so mu Ham so 
  
    
      
        y
        =
        f
        (
        x
        )
        =
        
          a
          
            x
          
        
      
    
    {\displaystyle y=f(x)=a^{x}}
  
 voi a la so thuc duong khac 1 đuoc goi la ham so mu voi co so a.


= Đao ham =
Ham so 
  
    
      
        y
        =
        f
        (
        x
        )
        =
        
          a
          
            x
          
        
      
    
    {\displaystyle y=f(x)=a^{x}}
  
 voi a la so thuc duong khac 1 thi co đao ham tai moi x va 
  
    
      
        
          y
          ′
        
        =
        
          a
          
            x
          
        
        ln
        ⁡
        (
        a
        )
      
    
    {\displaystyle y'=a^{x}\ln(a)}
  
 la đao ham cap 1 cua 
  
    
      
        f
        (
        x
        )
      
    
    {\displaystyle f(x)}
  

Đac biet ham so 
  
    
      
        y
        =
        
          e
          
            x
          
        
      
    
    {\displaystyle y=e^{x}}
  
 co đao ham cap 1 la 
  
    
      
        
          y
          ′
        
        =
        
          e
          
            x
          
        
      
    
    {\displaystyle y'=e^{x}}
  


= Chieu bien thien =
Ham so 
  
    
      
        y
        =
        f
        (
        x
        )
        =
        
          a
          
            x
          
        
      
    
    {\displaystyle y=f(x)=a^{x}}
  
 đong bien tren R neu a>1 va nghich bien tren R neu 0<a<1.


= Đo thi =

Đo thi ham so 
  
    
      
        y
        =
        f
        (
        x
        )
        =
        
          a
          
            x
          
        
      
    
    {\displaystyle y=f(x)=a^{x}}
  
co nhung tinh chat sau:

Luon đi qua điem I(0;1) va điem J(1;a)
Đo thi nam phia tren truc Ox va nhan truc Ox lam tiem can ngang


Tim chu so tan cung = Tim chu so tan cung cua luy thua =
Đe tim chu so tan cung, ta co the lap bang đe biet chu so tan cung đuoc thay đoi nhu the nao.
Vi du: Tim chu so tan cung cua 72004
Phan tich:

Giai:
Chu so tan cung đuoc lap lai theo day: 7, 9, 3, 1, 7,... (gom day 4 so hang lap lai)
2004 : 4 = 501 du 0
Vay chu so tan cung cua 72004 la 1.
(noi cach khac: 72004 = (74)501; vi 74 tan cung bang 1 nen (74)501 tan cung bang 1)


= Tim so cac so 0 tan cung cua mot tich =
Vi 2 × 5 = 10 nen muon tim so cac so 0 tan cung ta co the phan tich tich ban đau ra thua so nguyen to, tim so cap thua so {2, 5} la ra luon so cac so 0 tan cung.
Vi du: So 12! (12 giai thua) bao gom bao nhieu chu so 0 tan cung? 
Giai:
Ta co: 12! = 1 × 2 × 3 × ... × 12
Phan tich ra thua so nguyen to: 12! = 210 × 35 × 52 × 7 × 11
Vi co 10 thua so 2 va 2 thua so 5 nen tao đuoc 2 cap {2, 5}. 
Vay 12! co 2 chu so 0 tan cung.


Xem them Phep cong
Phep tru
Phep nhan
Phep chia
Phep khai can
Logarit
Vi phan
Gioi han
Tich phan
Tetration
He thap phan


Tham khao Thu muc Tran Van Hao va đong nghiep, Giai tich 12, Nha xuat ban giao duc