$\left(\frac{3}{4}x-\frac{2}{5}y\right)\left(\frac{3}{4}x+\frac{2}{5}y\right)$
$\left(3x+3\right)\left(3x-3\right)$
$4y^5-25y^3=0$
$1\:+2x\:-\:1$
$\frac{d}{dx}x^2-16$
$\frac{6x^2+15x+7}{x+2}$
$7\left(4-u\right)$
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