$\frac{x^3-3x-4x-12}{x^2-4}$
$17-3x\ge2$
$\sqrt[3]{\frac{x}{4x+1}}$
$3y^2dx=6x^4dy$
$\sqrt{\sec^2\left(x\right)-1}+\sqrt{\csc^2\left(x\right)-1}$
$\left(-4\right)-\left(-2\right)-\left(+3\right)-\left(6\right)$
$9\left(4-7\right)$
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