$4x\left(2+3x\right)-4+3x$
$3y^2+5y-2$
$\lim_{x\to\infty}\left(\frac{5t^2+3}{2t^3-t^2}\right)$
$2x^2-7x=4$
$= \frac { 5 } { 4 x - 12 }$
$\left(5+3.2\right)^2-4^2.2-3+\sqrt{2}^3-3.2-1$
$\left(-172\right)+\left(25\right)+\left(-108\right)$
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