$4\ln\left(x+6\right)=12$
$\frac{\left(x^4+x^3-11x^2+11x-3\right)}{x-3}$
$1-4x>2-x$
$\left(-8x^{6a}+12\right)\left(-8x^{6a}+15\right)$
$\frac{1}{4}x+\:2+\frac{2}{4}x-3$
$\sin\left(\frac{1}{\cos\left(x\right)}\right)$
$\left(y^{-1}+x^3\right)\left(y^{-1}-x^3\right)$
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