$\left(4a-3\right)\left(3a+2\right)$
$\lim_{x\to\infty}\left(\frac{4^x}{x}\right)$
$a+b\le c+d$
$\left(3x-2\right)\left(x\right)$
$\frac{4-2x}{x^2-4x^3}$
$3vu^2.\:-3v^2$
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