$\left(4y\:+\:5z\right)^2$
$\int\left(\frac{\left(x-2\right)}{\left(3x+4\right)^2}\right)dx$
$6x^2\:+\:4x\:-3y\:+31=0$
$\frac{dy}{dx}=36x$
$2x^2-x-3=0$
$10y^2-xy^2=x^3$
$\frac{6x^2y}{12xy^2}$
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