$\frac{d}{dx}e^{x^2}-y$
$4x^2-8x=1$
$\lim_{x\to\infty}\left(\frac{8}{e^x+6}\right)$
$\frac{\cos\left(u\right)+1}{\tan^2\left(u\right)}=\frac{\cos\left(u\right)}{\sec\left(u\right)-1}$
$\frac{3x^2-19x+25}{3x-4}$
$4.x+3=39$
$\left(\frac{1}{2x}+5x+10\right)^2$
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