$-x^2-3x+2;\:x=-2$
$\frac{4x-8}{3}<x+5$
$36xb-12xy$
$\int_{-\infty}^0\left(\frac{x+2}{\left(x^2+4x+4\right)}\right)dx$
$\left(4n^5+5n^6\right)^2$
$\frac{x^3+6x^2+5x+4}{x^2-3x+1}$
$\left(+8\right)-\left(-5\right)\cdot\left(4-7\right)$
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