$\frac{2x+3}{x}=\frac{x}{4}$
$\left(y+2\right)dx+\left(x+y\right)dy=0$
$x^2-9x=2$
$\frac{x}{x+3}+\frac{1}{x}=\frac{3}{x}$
$\left(a-15\right)\left(a+12\right)$
$\frac{-6}{x+5}\ge1$
$\left(a^3+2bc^2\right)^2$
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