$tan\left(x\right)=0,955$
$\left(\frac{2}{3}x^2-3\right)^2$
$\frac{m^2}{m^2+2m+1}+\frac{1}{3m+1}-\frac{1}{6}$
$x^2-8x+11\le0$
$\int-4tan^5\left(5x\right)dx$
$\left(4n+6\right)^3$
$\lim_{x\to\infty}\left(\frac{98x^3-55x-1.5}{x^3+.8x+.05}\right)$
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