$\frac{\left(\left(3x^2\right)-2x\right)}{x^2-1}>=4$
$\int x^3\left(4+x^2\right)dx$
$6\cos^{2}\theta-7\sin x-1=0$
$3x^2-12x=3$
$\lim_{x\to2}\left(\left(x-4\right)\left(x-2\right)\right)$
$-5c=40$
$-ln\left(cos\left(y\right)\right)=\frac{1}{2}ln\left(x^2+1\right)+c$
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