Datasets:

ddrg

/

math_formulas

like 4

-a^2 + b * b = (b - a) (b + a)

(y + 1)\cdot \left(y + 2\right)\cdot (y + 3)\cdot (y + 4) = (y^2 + 5\cdot y + 5 + 1)\cdot (y \cdot y + 5\cdot y + 5 + \left(-1\right)) = (y^2 + 5\cdot y + 5)^2 - 1 \cdot 1

\tfrac{1}{4}\cdot \pi + \frac{\pi}{4} = \pi/2

r \cdot r + r\cdot 3 + 2 = (r + 2)\cdot \left(1 + r\right)

x\cdot 2\cdot d = (x + d)\cdot (x + d) - x^2 + d^2

1 = c\cdot k^2\cdot 3 \Rightarrow c = \frac{1}{3\cdot k \cdot k}

z + k + 1 - x = k + 1 - x - z

(1 + 9)^{75} + (-1) = 10^{75} + (-1)

\tfrac{1}{A*B} = \dfrac{1}{B*A}

0 = 2*x \Rightarrow 0 = x

\binom{l}{t} = \frac1t*\left(l - t + 1\right)*\binom{l}{(-1) + t}

t < 1 + u,u < 1 + t\Longrightarrow t = u

p + q + 2\cdot (-1) = q + (-1) + p + (-1)

g + b + h = g + b + h

\lim_{r \to 0} \tan(2 \cdot r - \pi)/r = \lim_{r \to 0} \tan(2 \cdot r)/r = \lim_{r \to 0} \frac{\sin(2 \cdot r)}{r \cdot \cos(2 \cdot r)} = \lim_{r \to 0} \frac{\sin(2 \cdot r)}{2 \cdot r} \cdot \lim_{r \to 0} \frac{1}{\cos(2 \cdot r)} \cdot 2

\sum_{l=3}^{202} l = \sum_{l=5}^{204} (2*(-1) + l)

4 \cdot 4 \cdot 4 = 3^3 + 2\cdot 4^2 + 3^2 - 4 = 27 + 32 + 9 + 4\cdot (-1) = 64

11 = (5 + 1 + 2*(-1) + 3 + 4*(-1))*6 + 7*(-1)

23 = b + a \Rightarrow -a + 23 = b

(g + x) * (g + x) * (g + x) = g^3 + 3g^2 x + 3gx * x + x^3

2^3 + 1^3 = 3^2 \neq 3^3

2^{n \cdot n - n} = 4^{{n \choose 2}}

\sin(3 + z*2) = \sin(2*z + 3)

-12 \cdot i + 6 + 0 \cdot \left(-1\right) = -12 \cdot i + 6

333 = 3^5 + 3^2 + 3^4

-\dfrac{1}{64} \cdot 81 = 3/4 \cdot (-\frac{27}{16})

4z + 2(-1) = 7*(3z + 5) = 21 z + 35

(21 + 30 + 10 + 12 + 3)/14 = \frac{1}{7}\cdot 38

\sum_{i=1}^n (1 - (-1) + 1) = \sum_{i=1}^n (1 + 0\cdot (-1))

17 = (4 - j)\cdot (j + 4)

\sin(f - \frac{3*\pi}{2}) = \sin(-(3*\pi/2 - f)) = -\sin(3*\pi/2 - f) = \cos\left(f\right) = \cos\left(f\right)

11\cdot (9 + y) = y + 144 \Rightarrow y\cdot 11 + 99 = y + 144

\left(z\cdot 8 + z = 45 \Rightarrow z\cdot 9 = 45\right) \Rightarrow 5 = z

z + 2 \cdot (-1) = z + 2 \cdot \left(-1\right)

\sin{z} - z = \frac{1}{2 \cdot i} \cdot (e^{i \cdot x + R_n} - e^{-i \cdot x - R_n}) - z = \frac{1}{2 \cdot i} \cdot e^{R_n + i \cdot x} \cdot \left(1 - 2 \cdot i \cdot z \cdot e^{-(R_n + i \cdot x)} - e^{-2 \cdot (R_n + i \cdot x)}\right)

0 = 1 - \mathbb{P}\left(A\right) + \mathbb{P}(F) - \mathbb{P}(A)\cdot \mathbb{P}(F) = (1 - \mathbb{P}(A))\cdot \left(1 - \mathbb{P}(F)\right)

12 = \left(3^{1/2} \cdot 2\right)^2

d + d = (d + d)^2 = d^2 + d \cdot d + d^2 + d^2 = d + d + d + d \implies d + d = 0

\left(m + 1\right)^4/4 = \dfrac14 \cdot (1 + m^4 + 4 \cdot m^3 + m^2 \cdot 6 + 4 \cdot m)

\tfrac{1}{12} \cdot 3 \cdot \dfrac{4}{13} = \frac{1}{13}

(\frac{1 + s}{1 - s})^{1 / 2} = \dfrac{1 + s}{(1 - s^2)^{\frac{1}{2}}} = (1 + s) \cdot E[s]

g^f \cdot g^f = g^{2 \cdot f}

960 = 4\cdot 4\cdot 6\cdot 5\cdot 4/2

853 = 29 \cdot 29 + 3^2 + 1 + 1 + 1

\frac{1}{3}\cdot 20 = 4\cdot 5/3

\dfrac37 = \frac{1}{21} \cdot 9

k = d + x \cdot r \implies r \cdot x = k - d

7^{2 \cdot (k + 1) + 1} + 1 = 7^{2 \cdot k + 1 + 2} + 1 = 49 \cdot \left(7^{2 \cdot k + 1} + 1\right) - 8 \cdot 6

-y = x^2 - 2 \cdot x + 7 \cdot (-1) = \left(x + (-1)\right)^2 + 8 \cdot (-1)\Longrightarrow -(y + 8 \cdot (-1)) = ((-1) + x)^2

-\dfrac12 = 1 + 3 + 9 + 27 + ...

x\cdot (z + z) = l_x\cdot \left(z + z\right) + l_y\cdot (z + z) = l_x\cdot z + l_x\cdot z + l_y\cdot z + l_y\cdot z = l_x\cdot z + x\cdot z + l_y\cdot z

-2^3 + 3 \times 3 = 1

45 = (2 + 1) \cdot \left(1 + 4\right) \cdot (2 + 1)

\tfrac{4}{(3 + z)\times (6\times (-1) + z)} = \dfrac{4}{z^2 - z\times 3 + 18\times (-1)}

-\frac{1000}{-200} = -\frac{1}{-40} \cdot 200 = -40/(-8) = 5

\dfrac{1/2 + 2/4 + \frac18*4 + \cdots + \frac{1}{2^x}*2^{x + (-1)}}{\dfrac12 + 1/4 + \frac18 + \cdots + \frac{1}{2^x}} = x/2*\frac{1}{2^x + (-1)}*2^x = \frac{x*2^{x + (-1)}}{2^x + (-1)}*1

\sin(\pi/8)\cdot \sin(3\cdot \pi/8) = \frac{1}{2}\cdot \left(\cos(\frac{1}{8}\cdot \pi - 3\cdot \pi/8) - \cos(\pi/8 + 3\cdot \pi/8)\right) = (\cos\left(\frac14\cdot ((-1)\cdot \pi)\right) - \cos\left(\pi/2\right))/2 = \sqrt{2}/4

\dfrac12 (\pi - \dfrac{\pi}{6}) = 5 \pi/12

\frac{(-5) \cdot \frac{1}{4}}{3 \cdot 1/5} = -\frac{5}{4} \cdot 5/3

\left(B*3 = \frac{1}{B} \implies I = B * B*3\right) \implies I/3 = B^2

-3 = -3/1

3 = 24 + 5\cdot z + 30\cdot \left(-1\right) = 5\cdot z + 6\cdot \left(-1\right)

a_x \cdot x + E_{\nu} = -(a_x \cdot x + E_{\nu}) = -a_x \cdot x - E_{\nu}

5^3=5^2\cdot 5

\frac{-i \cdot 13 - 18}{-i - 4} = \frac{-4 + i}{-4 + i} \cdot \frac{1}{-4 - i} \cdot (-18 - i \cdot 13)

S\cdot A = A\cdot S

\frac{\text{d}}{\text{d}x} (x \cdot x)^{1/2} = \frac{2}{(x^2)^{1/2} \cdot 2} \cdot x

0 = z^2 - \frac56\times z + 1/6 = (z - \dfrac12)\times \left(z - 1/3\right)

(X + (-1)) (X^2 + X + 1) = \left(-1\right) + X^3

-(15\cdot k + 27\cdot (-1))/(60\cdot k) = \frac{3}{3}\cdot (-\dfrac{1}{20\cdot k}\cdot (5\cdot k + 9\cdot \left(-1\right)))

\frac{1}{z} = \dfrac1z*\overline{z}/\overline{z} = \dfrac{\overline{z}}{|z|^2}

\sigma^2 + \sigma \cdot 4 + 1 = 0 \implies \sigma = -2 \pm \sqrt{3}

\dfrac12\cdot e^{y\cdot 2} = \frac{1}{2}\cdot x^2 - 8\cdot x + G \Rightarrow e^{2\cdot y} = 2\cdot G + x^2 - x\cdot 16

4\cdot 1^{-1}/5 = \frac45\cdot 1^{-1}

3 \cdot x = \dfrac{3}{x^2} \cdot x^2 \cdot x

n\cdot 2^{n + 1} + 2^{n + 1} = (n + 1)\cdot 2^{1 + n}

-hd = h\cdot (-d)

3 \cdot 11^{1/2} = (2 + 1) \cdot 11^{1/2}

\left(Z + x\right)^2 = x^2 + Z^2 + 2Zx

3\left(z_1 + z_2\right) = 3z_2 + z_1*3

5\cdot \varepsilon^2 + 320\cdot (-1) = 5\cdot (\varepsilon^2 + 64\cdot (-1)) = 5\cdot (\varepsilon + 8)\cdot (\varepsilon + 8\cdot (-1))

\ln(-5 \cdot t + 2000) = \ln(5) + \ln(400 - t)

130/117 = \frac{130}{9 \cdot 13} \cdot 1

\frac{1}{2^{(-1) + m}} = \dfrac{1}{2^m\cdot \frac{1}{2}}

5 + a^2 = 5 + g^2\Longrightarrow g^2 = a^2

\frac{a^5}{a^4}\cdot 3/2 = \frac{a^5\cdot 3}{a^4\cdot 2}\cdot 1

\pi = -3/4*\pi + \pi*\frac{7}{4}

-\theta_1 + \theta_2 \cdot 2 = k \cdot \pi \cdot 2 + \theta_1 \Rightarrow \theta_2 = \theta_1 + \pi \cdot k

d^2*4 + 4x^2 + 64 - 4dx - 16 d - x*16 = 3\left(-x + d\right)^2 + (d + x + 8(-1))^2

12 = 3 + 3 + 6

11/19\cdot \dfrac{1}{20}12 = \dfrac{1}{380}132

h = h \cdot b = b\Longrightarrow h = b

\sum_{n=1}^\infty \frac{1}{n^4} = \sum_{n=1}^\infty \frac{1}{(n*2)^4} + \sum_{n=0}^\infty \frac{1}{(1 + n*2)^4} rightarrow \sum_{n=0}^\infty \dfrac{\dots}{(1 + n*2)^4} = (\sum_{n=1}^\infty \dfrac{1}{n^4})*\frac{15}{16}

\frac{40}{28} = \frac{10}{7}

\dfrac{2}{\pi} = 4 \cdot \frac{1}{\pi}/2

\frac{1}{12}\cdot 60 = 5

2 = 10^{1 / 2} (a + b*10^{1 / 2}) = a*10^{1 / 2} + 10 b

74^{\frac{1}{2}} = (8 \cdot (-1) + 74^{\tfrac{1}{2}})/1 + 8

5 = 7479 - 202\cdot 37

4/10 \cdot 9 - \frac{1}{10} \cdot 6 \cdot 6 = 0