$x^2-10x+25=1$
$\lim_{x\to1}\left(\frac{ln\left(4-3x\right)}{x-1}\right)$
$\frac{d}{dx}\left(2e^{\tan\left(x\right)}\right)$
$x^2+11x+28=0$
$3x-1+6x-2$
$2x^2\:-\:3x\:+\:2,\:para\:x=-1$
$3\cdot2-1-6$
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