$\frac{6}{x+1}=\frac{5}{x-1}$
$6x^2+31.2x+15.56$
$\int\frac{4}{\sqrt{4-u^2}}du$
$\left(-100\right)\:\left(-80\right)$
$-2+\left(4-5+2\right)-3-1-\left|6+\left(-3-1\right)-\left(-2+4\right)\right|+3-4$
$3^{x-2}=9$
$k^3+p^3$
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