$\left(x+1\right)\left(x+12\right)$
$\left(-xy^2\right)\cdot\left(3x^2y\right)$
$2\left(v+4\right)-4v$
$7,8\cdot4$
$\frac{x+4}{x^2+8x+12}$
$\left(3x-5y^2-6\right).\left(2x+3y^2\right)$
$\frac{sec^2\left(x\right)}{1+cot^2\left(x\right)}=tan^2x$
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