Simplifying
$16\le4\cdot12<60$
$\int\frac{z-1}{3z+1}dz$
$\frac{x^3-2x^2+x+1}{x-1}$
$x^2+x-11\cdot4x-16$
$y=\frac{\sqrt[2]{\left(2x+1\right)\cdot\left(3x+2\right)}}{4x+3}$
$-4h+8+9h-1+3h-7$
$\left(\sec m-1\right)\left(\sec m+1\right)=\tan^2m$
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