$\left(4+4\right)\left|-1\right|$
$3x-2y+5y-4x-3x+y$
$5x+16<8x-5$
$\int\frac{x^{3\:}-2\sqrt{x}}{x}dx$
$\lim_{x\to4}\left(\frac{3-\left(x-4\right)}{x^2-16}\right)$
$6x+6=48$
$\int\left(2f^2-1\right)^2dx$
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