$x-2<5$
$x^2-4x-17\le4$
$6\left(x+1\right)-4x=5x-9$
$-44-8x$
$\left(2x^3-x^2\right)^2$
$\tan\left(x\right)=-12$
$\frac{tan\left(x\right)}{sin\left(2x\right)}=\:\frac{1}{2\left(1\:+\:cot^2\left(x\right)\right)}$
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