$4-3x\le-1-8x$
$\left(12x^6y^4\right)\::\:\left(3x^4y^3\right)$
$14r+7-3r$
$\int xsec^2\left(10-x^2\right)dx$
$\frac { x ^ { 2 } - 8 x - 6 } { 3 ( x + 1 ) ( x ^ { 2 } + 5 ) }$
$-10h+22h-7c-10c+8r-22r$
$5\sqrt{x^3}$
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