$\frac{x^4-x^3+4x+1}{x^2+3}$
$5x-4<12$
$y'+xy=xy^2$
$\left(7x\:-\:5\right)\:\left(7x\:+\:8\right)\:$
$5\left(4x-7y+12\right)$
$\frac{2x-1}{x+2}=-3$
$3+2y+3x+3$
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