$10x-6<8x+4$
$\frac{x^2-4}{x+4}$
$\left(8+-10\right)\cdot\left(-7\right)^2$
$7+\frac{a}{3}=-5$
$dx\:\left[\left(4x^2+7\right)^2\left(2x^3+1\right)^4\right]$
$\frac{1-\tan^2a}{1+\tan^2a}+1=\frac{2}{\sec^2a}$
$3x^2+18=0$
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