$x^2-4x+5$
$-3\cdot6^2$
$\frac{564}{2}$
$\frac{\sin\:^2\left(x\right)-1}{\sin\:\left(x\right)+1}$
$2\cdot\left(3+1\right)+2$
$\lim_{x\to\infty}\left(\sqrt[4]{16x^4-x^3}-2x\right)$
$4x^2+4x+4-5x^4-3x^3+2x^2-x+3$
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