$8-\left(-4-2+3\right)+1$
$x^2+4x-6=0$
$\lim_{x\to\infty}\left(\frac{x+3}{3}\right)^{4x+1}$
$\:2x\:+\:1.\:para\:x=-2$
$\int\frac{8u^2}{\left(1+u^2\right)^3}du$
$\frac{d}{dx}8^{x-3}$
$4x+10=5\left(x-2\right)$
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