$x^2+3y\frac{dy}{dx}=0$
$\frac{3}{a+2}+\frac{8}{a-5}$
$\left(2a+b\right)^2-\left(2a-b\right)^2+8ab$
$\left(1\right)^3-2\left(1\right)^2-3\left(1\right)+1$
$-13-3\cdot2$
$4=\frac{b}{3}$
$2x^2-x-4$
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