$\int\frac{4x+6}{x^4-x^2}dx$
$\left[\:7+\left(-19\right)-\left(-8\right)\:\right]+\:\left(-1\right)\:$
$\left(z-10\right)\left(z+15\right)$
$6rs^2-18rs+42rs^2s^2$
$0.\left(-17\right)=-17$
$2x-3-4\left(x^2-1\right)\ge10+x-4x^2$
$\frac{3x+2}{8}\geq0$
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