$\frac{\left(2x^2y^{-3}\right)}{\left(x^2y^{-1}\right)^{-2}}$
$t^2-8t+16$
$12.89+96.54+59.80$
$\frac{x}{x-3}-2=\frac{1}{x-3}$
$40+\left|25-\left(3+2\right)\right|$
$4x^2+x+8$
$2\left(x+1\right)+3>5\left(x-2\right)+7$
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