$\left(\frac{\left(\tan^2\left(x\right)\right)\left(\csc\left(x\right)\right)}{\left(x^2+4x\right)}\right)$
$-10x+-5x+3z$
$\frac{1}{4}y^6+y^{-6}$
$-56+-1$
$x^2-8x\le0$
$y=\cos\left(3\right)x$
$123=3$
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