$-19\:-\:7$
$\left(4x-6\right)\left(4x+1\right)$
$cos3x\:=\:-1$
$\lim_{x\to+0}\left(x^3\ln\left(x\right)\right)$
$m^2-4m-12$
$\sqrt[3]{3x-8}=4$
$-12+6x\:-2x+7\:$
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