$\lim\:_{x\to\:\:0}\left(\frac{t^2+t}{3}\right)$
$\left(7\left(8-4\right)+16\right)-\left(15-\left(7+3\right)\right)$
$xdu=\frac{x-2ux}{x}dx$
$x^4+x^3+6x$
$5-1^4+2-1+2^2+3+5-1^3-4$
$5x^2-3x+1$
$\left(4a+2b\right)\left(4a-2b\right)$
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