$5a+3-6b-1-3a$
$\lim_{x\to a}\left(\frac{x+1-e^x}{x^2}\right)$
$4\left(6x+12\right)$
$\left(x+5\right)^2-3x=\left(x+3\right)^2+3x-2$
$\lim_{x\to\infty}\left(\left|\frac{\left(x+1\right)^2}{ex^2}\right|\right)$
$\left(4a^2-5b^3\right)^2$
$7.6\:-\:5.88\:=\:\:\:$
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