$\int\sec^{12}\left(x\right)\tan\left(x\right)dx$
$2a^2-8ab+8b^2$
$9\sin^2x-12\sin x=0$
$12x-7<18x-9$
$\left(\frac{1}{3}\right)^2\left(-\frac{1}{3}\right)^{-2}$
$\left(6x^2-7x\right)^2\left(4x-2\right)^3$
$x\cdot2-6=x-6$
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