$\left(10x^4-3x^2+15\right)\cdot\left(2x+3\right)$
$\int\tan\left(4-9x\right).dx$
$\frac{2x-5}{4}\le3$
$\frac{d}{dx}\frac{x+y}{x-y}=0$
$\frac{4x-16}{5x+5}$
$3a+6a+9a$
$9x^2-12x-5$
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