$2ydx\:+3xdy=0$
$\left(x^3y^5z^4\right)^3$
$\frac{-\sec^2x}{\left(\sec^2x\right)\cdot\tan\left(y\right)}$
$1y-4y$
$x+2y-42=-4$
$3.141592653589793=\pi$
$\left(3-u\right)^3$
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