$\left(4y-7\right)\left(4y-2\right)$
$x^2-x-10;\:x=5$
$\frac{2}{x}=\frac{1}{x-2}-3$
$\frac{x^5}{\left(1+x^2\right)^2}$
$\left|1+\left(-83\right)\right|-37$
$x^2\:+\:x\:1$
$\left(a+32\right)^2$
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