$2a+6b^2$
$3^2>2^3$
$\frac{d}{dx}\:\frac{x^3-4x^2+2}{\sqrt[3]{x}}$
$-7-17.68$
$6x-6=24$
$\sqrt[4]{b^{16}}$
$\left(f\cdot g\right)\left(\frac{3}{2x-1}\right)\left(\frac{2}{x}\right)$
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