$\frac{10^8}{10^2}$
$\lim_{x\to\infty}\:\frac{\left(3x^2-4\right)}{3-7x}$
$4-8\ln\left(x\right)=9$
$\log12+2\log\left(x\right)$
$4+2+6$
$\left(-3\right)-\left(-16\right)\left(+8\right)$
$\left(10x^2+3y^3\right).\left(10x^2-3y^3\right)$
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