$32^{x-1}=16^{x+4}$
$\tan\left(a\right)+1=\sin\left(a\right)\cdot\sec\left(a\right)+1$
$\left(2x^2\:+\:5x\right)\:\left(x^3\:+\:7x^2\:-\:3x\:+\:4\right)$
$\left(3x^4-6y^3-9z^2\right)^2$
$x^2+4x+3>0$
$-4\left(-7r-1\right)+5r$
$\int\frac{1}{5x^2-2x+5}dx$
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