$12.8\cdot40+3\left(x\right)$
$\frac{h-13}{9}=-2$
$-\left(x-6\right)^2$
$g\left(x\right)=\left(x^2+1\right)^2\left(x-1\right)^5x^3$
$12+4-16+48-3$
$\:y'\:+\:y^2\:-\:y\:=\:0$
$\left(5-7\right)^4-\left(-11\right)\cdot\left(-3\right)$
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