$3\left(x-4\right)\le6$
$\frac{14x+7}{x^2+x-6}$
$\frac{1}{6-x},\:c=1$
$\lim_{x\to0}\left(\frac{\ln\left(4\cdot e^x-4\right)}{\ln\left(x\right)}\right)$
$5=4z+13$
$\left(2x\sqrt{4-3x}\right)\left(2\sqrt{4-3x}\right)$
$-32x^2+18\cdot x-17=0$
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