$\frac{d}{dx}16x^3+9x^2y=54x$
$\frac{1-\frac{1}{\sec^2\left(x\right)}}{\sec\left(x\right)\sin^2\left(x\right)}$
$2x+\:3<9$
$\left(-7x-3\right)\left(-11+2x\right)$
$x^3+11x^2+10x=0$
$\int x^2.\sqrt[6]{4+x}dx$
$4x^2-11^2-3$
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