$y'-xy=1-x$
$-2\left(x+y\right)^5$
$16x^2-9y^2=144$
$\frac{1}{5}-\frac{1}{5}\cdot5+\frac{1}{5}-3^{2}-\left(\frac{1}{5}-1+\frac{1}{5}-9\right)$
$\frac{b}{6}-1>1$
$\frac{1}{4}\tan\left(x\right)$
$26x^2+x+4$
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