$x^2-18x=40$
$y'\:=\:-3y+8+4e^{-4s}$
$\left(-2\right)\cdot\left(-2\right)\cdot2\cdot\left(-3\right)$
$6y^2+48y$
$\int\log\left(xe^{7x}\right)dx$
$\int x ( x ^ { 4 } - 1 ) d x$
$2x-3\ge\frac{2}{3}$
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