$y'=\frac{coty}{t}$
$\left(3x^2+u^2\right)x'+2xy+3u^2$
$\frac{\left(4x^3-2x^2+2x+1\right)}{\left(2x+1\right)}$
$\frac{1}{2}\left(w^{-4}\right)^2\cdot\frac{12}{2}\left(w^3\right)^3$
$3\left(5x\:-\:6\right)\:-\:2\left(4x\:+\:5\right)$
$\frac{\left(3x-1\right)}{x^2+1}\le1$
$-5x^6\cdot2x^2$
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