$\frac{x^2}{x+6}$
$\frac{1}{1-\sin^2a}=1+\tan\left(a\right)$
$\frac{1+\tan^2\left(\infty\right)}{1-\tan^2\left(\infty\right)}$
$25n^2-16m^2$
$-x^2-11-2a^3b^4+6a^3b^4-11y^3-5x^2$
$\int e^{5x}\cdot sin\left(bx\right)dx$
$2x+10>18$
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