$\left(4a^3b^5\right).\left(-\frac{1}{4}abc^2\right)$
$12\le6-2x$
$\lim_{x\to0}\left(\frac{ln\left(1+x\right)-x}{tan^2x}\right)$
$\left(-59\right)-\left(-18\right)$
$2y^2+4y^2-8y-16$
$2^{10.3}\cdot0.7$
$\lim_{x\to0}\left(\frac{x}{-x-1}\right)$
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