$sec^2a\cdot tan^2a$
$2x^2-3x+6$
$x-3\le12$
$\left(\frac{a^5b^4}{x^6b^3}\right)^0$
$x^2-8x+16=64$
$\lim_{x\to h}\:\:\frac{x^2+h^2}{x+h}$
$3-2\cdot\left(3\cdot2-4\cdot2\right)+\left(-3\right)\cdot\left(2-5\right)$
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