$-12\cdot14$
$\frac{1}{\sqrt{16x^2+47}}$
$\lim_{x\to3}\left(\frac{\sqrt{3}-\sqrt{x}}{x^2-9}\right)$
$0,2\cdot0,2$
$\:\:11x\:-\:6\:=\:3x\:+\:26$
$\left(x+5\right)\left(x-5\right)>0$
$6-\sqrt{2x-1}\sqrt{x+4}$
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