$\lim_{x\to0}\left(\frac{-sin\left(x\right)}{1+sin\left(x\right)}\right)$
$\left(4+6y\right)^3$
$5x^2+5ay^2$
$xy-3y=x^5$
$3x-4\le x+8$
$10x-18$
$-4in\left(x+1\right)\sin\left(4x\right)+\frac{cos\left(4x\right)}{x+1}$
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