$6-5x<8+3x$
$\left(3\sqrt{5}-2\sqrt{7}x\right)\left(3\sqrt{5}+2\sqrt{7}\right)$
$6\cos^2x+3\sin x-3$
$\sqrt{\left(4-0\right)}^2\:+\:\left(64-0\right)^2$
$\frac{1}{x}+\frac{12}{x-3}=-4$
$\frac{2}{x+1}=\frac{3}{x-1}1$
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