$\int\left(\frac{1}{\left(10x-3\right)^2}\right)dx$
$-\left(-5\right)^2+6\left(-5\right)$
$\sqrt[4]{x^{12}}y^8$
$\left(\left(5x^2+1\right)^3\left(7x-3\right)^4\right)$
$\sqrt{x-14}=9$
$3x+2>x+12$
$x^2-6x\ge-8$
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