$2\sin^2x+2=5\sin x$
$x\left(x+1\right)^3$
$5\:x\:5\:x\:5\:x\:5\:x\:5$
$\left(5x\:+\:4\right)\:\left(5x\:+\:8\right)$
$\lim_{x\to\infty}\:\frac{x^2+2}{x^2+1}$
$14x^4y+7xy^2+21$
$\frac{1}{\sqrt{2x^2+3x+1}}$
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