$-\frac{1}{e^{\left(\infty\right)^2}}$
$x-5=13$
$6\left(x-3\right)^2+7\left(x+3\right)-3$
$\:a^3x^{-5}z^5$
$12c^2d-24cd^2+6c^3d^2$
$\left(5x^4-6x^3+x\right)$
$\int\left(-\frac{7x}{6-3x^2}\right)dx$
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