$\left(\frac{3}{4}m+\frac{2}{3}n\right)^2$
$\left(x^2-3y^4\right)^2$
$3x-36$
$21x-40=\frac{21}{x}$
$\frac{3x^2+16x+21}{x+4}$
$\frac{6x^4y}{16y^3}$
$dz\:-\:2te^z\cdot dt=0$
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