分母有理化

分母有理化，简称有理化，指的是将该原为无理数的分母化为有理数的过程，也就是将分母中的根号化去。 有理化后通常方便运算，有理化的过程可能會影響分子，但分子及分母的比例不變。 应用一般根号运算： 1 a = 1 a a a = a a {\displaystyle {\frac {1}{\sqrt {a}}}={\frac {1{\sqrt {a}}}{{\sqrt {a}}{\sqrt {a}}}}={\frac {\sqrt {a}}{a}}} 1 a n = a n − 1 n a {\displaystyle {\frac {1}{\sqrt[{n}]{a}}}={\frac {\sqrt[{n}]{a^{n-1}}}{a}}} 应用平方差公式： 1 a + b = a − b a − b {\displaystyle {\frac {1}{{\sqrt {a}}+{\sqrt {b}}}}={\frac {{\sqrt {a}}-{\sqrt {b}}}{a-b}}} 1 a − b = a + b a − b {\displaystyle {\frac {1}{{\sqrt {a}}-{\sqrt {b}}}}={\frac {{\sqrt {a}}+{\sqrt {b}}}{a-b}}} 1 a + b = a − b a − b 2 {\displaystyle {\frac {1}{{\sqrt {a}}+b}}={\frac {{\sqrt {a}}-b}{a-b^{2}}}} 1 a − b = a + b a − b 2 {\displaystyle {\frac {1}{{\sqrt {a}}-b}}={\frac {{\sqrt {a}}+b}{a-b^{2}}}} 应用立方和、立方差公式： 1 a 3 + b 3 = a 2 3 − a b 3 + b 2 3 a + b {\displaystyle {\frac {1}{{\sqrt[{3}]{a}}+{\sqrt[{3}]{b}}}}={\frac {{\sqrt[{3}]{a^{2}}}-{\sqrt[{3}]{ab}}+{\sqrt[{3}]{b^{2}}}}{a+b}}} 1 a 3 − b 3 = a 2 3 + a b 3 + b 2 3 a − b {\displaystyle {\frac {1}{{\sqrt[{3}]{a}}-{\sqrt[{3}]{b}}}}={\frac {{\sqrt[{3}]{a^{2}}}+{\sqrt[{3}]{ab}}+{\sqrt[{3}]{b^{2}}}}{a-b}}} 1 a 3 + b = a 2 3 − a 3 b + b 2 a + b 3 {\displaystyle {\frac {1}{{\sqrt[{3}]{a}}+b}}={\frac {{\sqrt[{3}]{a^{2}}}-{\sqrt[{3}]{a}}b+b^{2}}{a+b^{3}}}} 1 a 3 − b = a 2 3 + a 3 b + b 2 a − b 3 {\displaystyle {\frac {1}{{\sqrt[{3}]{a}}-b}}={\frac {{\sqrt[{3}]{a^{2}}}+{\sqrt[{3}]{a}}b+b^{2}}{a-b^{3}}}} 1 a + b + c = a − b − c a − b − c 2 − 2 c b {\displaystyle {\frac {1}{{\sqrt {a}}+{\sqrt {b}}+c}}={\frac {{\sqrt {a}}-{\sqrt {b}}-c}{a-b-c^{2}-2c{\sqrt {b}}}}} 设 x = 2 3 {\displaystyle x={\sqrt[{3}]{2}}} 有理化 1 1 + 2 2 3 + 3 4 3 {\displaystyle {\frac {1}{1+2{\sqrt[{3}]{2}}+3{\sqrt[{3}]{4}}}}} ( x 3 − 2 ) u ( x ) + ( 1 + 2 x + 3 x 2 ) v ( x ) = 1 {\displaystyle (x^{3}-2)u(x)+(1+2x+3x^{2})v(x)=1} u ( x ) = − 1 89 ( 50 + 3 x ) , v ( x ) = 1 89 ( − 11 + 16 x + x 2 ) {\displaystyle u(x)={\frac {-1}{89}}(50+3x),v(x)={\frac {1}{89}}(-11+16x+x^{2})} 1 1 + 2 2 3 + 3 4 3 = v ( 2 3 ) = 1 89 ( − 11 + 16 2 3 + 4 3 ) {\displaystyle {\frac {1}{1+2{\sqrt[{3}]{2}}+3{\sqrt[{3}]{4}}}}=v({\sqrt[{3}]{2}})={\frac {1}{89}}(-11+16{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}})} x 3 = 2 x 2 + 3 x + 4 {\displaystyle x^{3}=2x^{2}+3x+4} ，求 1 3 + 2 x + x 2 {\displaystyle {\frac {1}{3+2x+x^{2}}}} 设 ( 3 + 2 x + x 2 ) ( a + b x + c x 2 ) = 1 {\displaystyle (3+2x+x^{2})(a+bx+cx^{2})=1} ( 1 x x 2 x 3 x 4 ) ( 3 0 0 2 3 0 1 2 3 0 1 2 0 0 1 ) ( a b c ) = ( 1 x x 2 x 3 ) ( 3 0 0 2 3 4 1 2 6 0 1 4 ) ( a b c ) = ( 1 x x 2 ) ( 3 4 16 2 6 16 1 4 14 ) ( a b c ) = ( 1 x x 2 ) ( 1 0 0 ) {\displaystyle {\begin{pmatrix}1&x&x^{2}&x^{3}&x^{4}\end{pmatrix}}{\begin{pmatrix}3&0&0\\2&3&0\\1&2&3\\0&1&2\\0&0&1\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}={\begin{pmatrix}1&x&x^{2}&x^{3}\end{pmatrix}}{\begin{pmatrix}3&0&0\\2&3&4\\1&2&6\\0&1&4\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}={\begin{pmatrix}1&x&x^{2}\end{pmatrix}}{\begin{pmatrix}3&4&16\\2&6&16\\1&4&14\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}={\begin{pmatrix}1&x&x^{2}\end{pmatrix}}{\begin{pmatrix}1\\0\\0\end{pmatrix}}} ( a b c ) = ( 3 4 16 2 6 16 1 4 14 ) − 1 ( 1 0 0 ) = 1 22 ( 10 − 6 1 ) {\displaystyle {\begin{pmatrix}a\\b\\c\end{pmatrix}}={\begin{pmatrix}3&4&16\\2&6&16\\1&4&14\end{pmatrix}}^{-1}{\begin{pmatrix}1\\0\\0\end{pmatrix}}={\frac {1}{22}}{\begin{pmatrix}10\\-6\\1\end{pmatrix}}} 1 x 2 + 2 x + 3 = x 2 − 6 x + 10 22 {\displaystyle {\frac {1}{x^{2}+2x+3}}={\frac {x^{2}-6x+10}{22}}} 辗转相除法 数学主题 分母有理化与分子有理化. [2013-10-10]. （原始内容存档于2019-06-03）.  韩士安 林磊. 近世代数（第二版）.