$\sqrt[3]{\frac{x}{y}}$
$x^2-x>2\left(1-x\right)$
$2x^{2}+8x-6$
$3x-9\le2x+6$
$\int\frac{1}{\left(x^2\right)\left(\sqrt{16\cdot x^2-9}\right)}$
$2y^2y'=3e^{2x}$
$-3.\sqrt[2]{242}$
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