$\frac{x+5}{12}=\frac{2}{x}$
$13y^2+5x+y-6y^2-x$
$\frac{\left(-5^4\right)}{\left(-5^3\right)}$
$\left(3x+4y\right)\left(9x^2-12xy+16y^2\right)$
$2\cos^2x+5\cos x-3=0$
$\frac{x^3}{4}-3x^4y+9y^2$
$x\:-\:5\:=12$
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