$x^2y+4xy^2^2$
$9+\frac{5}{4}x\cdot10x-28$
$cos\:\left(\pi\:\:\:-\:x\right)+\:sin\:\left(\frac{\pi\:\:\:\:}{2}+\:x\:\right)=\:0$
$\frac{dy}{dx}\cdot\left(1+x\right)^2=\left(1+y\right)^2$
$x^2+10x=-50$
$2xy\:+\:2x$
$-4-\left(2.8\right)$
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