$\frac{1-\tan\left(x\right)^2}{1+\tan\left(x\right)^2}=\cos\left(x\right)^2+\sin\left(x\right)^2$
$\frac{4x-8}{3}<x+5$
$\left(8zc+150y\right)^2$
$\left(8x^2y-9m^3\right)^2$
$5x+3y+4x\:-2y$
$\lim_{x\to1}\left(\frac{x^2-1}{x+3}\right)$
$x^2-2x=-26$
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