$\frac{m^8}{m^3}$
$2,5\:+\:18,75$
$-2x^2+18\ge0$
$-2x\le-20$
$\int\left(40-\frac{2}{\sec\phi}\right)d\phi$
$1x^2-8x+17$
$\left(x^2+2x\right)\ln\left(x\right)$
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