$\left(3+10ab\right)\left(3-10ab\right)$
$\frac{\tan^2x}{\sec x}\left(\sec x-\cos x\right)$
$2j+1\le-1$
$-x^2+2x+3>0$
$\left(-39\right)\::\:\left(-13\right)\:$
$\frac{9x^2}{3x^3}$
$3\left(-2.7\right)+2\left(-2.7\right)-7+3\left(7\right)$
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