$\tan\left(2x\right)=\frac{4}{\cot\left(x\right)\tan\left(x\right)}$
$\left[\left(3\right)-\:\left(-2\right)\right]\:-5$
$3x^2-6x=3$
$\frac{16a^3b+8a^2b+24ab^2}{8a}$
$\left(d^4+2d^3-6d^2-16d-8\right)y=0$
$y'+e^{\left(-x\right)}\cdot y^2+y=0$
$\sqrt{xe^e}$
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