$\lim_{x\to1}\left(\frac{8x^2+4x-6}{4x-3}\right)$
$\frac{1}{2}ab-\frac{3}{4}ab$
$2\cos^2x-3\cos\left(x\right)=-1$
$25x^2-6x+36$
$\left(5+x\right)\frac{dy}{dx}=8+x$
$x^3-5x-2x^3$
$\frac{dx}{dt}=\frac{\left(8xlnx\right)}{t}$
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