$\frac{d}{dx}\:y=\left(3x^3-1\right)\left(2x+5\right)^3$
$\left(\left(2a-3b\right)^2+\left(3a-5b\right)^2\right)^2$
$-\frac{4}{3}x+11=\frac{x}{2}$
$7x+4x^2-5x$
$3\cot^2x+10\csc x=5$
$x^2-4x+144=0$
$-3x\:>\:6$
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