$\frac{\left(x-1\right)^2}{4}=\frac{x+5}{6}+x$
$\:\lim_{x\:\to\:0}\:\frac{1-\cos\:x}{1+x-e^x}$
$\frac{10}{x^2+5x-6}$
$\frac{x^2-6x+8}{x-2}$
$4\cdot\left(-1\right)$
$log3\left(7x-2\right)=5$
$1b$
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