$\left(x+y+n\right)^2$
$3x^2-6x+8=\left(x-2\right)x$
$x^2+8x+18=0$
$2\cdot5+3\cdot2$
$\lim_{x\to\infty}\left(\frac{\left(3x^2+2x\right)}{5x^3-3}\right)$
$\lim_{x\to\infty}\:\frac{x}{\sqrt{x^2-x}}$
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