$\frac{x^3+5}{x^2-1}$
$12,3\cdot12,746$
$\left(6x^3\right)\left(3x^{-5}\right)$
$\frac{x^{\frac{1}{4}}}{5x}$
$2xy\:dx\:+\:\left(x^2+\cos\left(y\right)\right)dy=0$
$\frac{\cos\left(y\right)e^{2x}}{2}-e^x\cos\left(y\right)$
$x+5>18$
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