$m^8+m^4-2$
$\frac{4a^4h^6}{60a^4h^2}$
$\lim_{x\to\infty}3x^{-2}e^x$
$\left(x^2+5z\right)^3$
$4x-x^2>12$
$40:\left(-5\right).3-\left(-6\right)$
$\left(13x-7\right)\left(17x+7\right)$
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