$x'+5x=\frac{1}{2}$
$\int_0^{\pi2}\left(\frac{1-\tan^3\left(x\right)}{\sec^2\left(x\right)}\right)dx$
$\sqrt[4]{81x^{40}}$
$2\left(2q+5\right)$
$2\sqrt{16^3}$
$a^2+10a+16$
$\frac{-a^2b^3c+\frac{1}{2}\:a^3b^2c^2-\frac{3}{4}a^4bc^3}{-\frac{6}{5}ac}$
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