$\left(3x+4\right)\left(x+4\right)$
$-16+m^4$
$\int_0^5\pi\:\ln\:^2\left(3x+1\right)dx$
$7,8\cdot4$
$20:\left(-4\right):\left(-5\right)$
$\left(x^{12}\right)^8$
$\frac{-1}{16\left(x-3\right)}$
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