$2\cdot\sqrt{12a}$

$1\frac{3}{4}+1\frac{1}{4}$

$x^2+20x-18=3$

$\cos\left(\arctan\left(\frac{12}{5}\right)+\arcsin\left(\frac{3}{5}\right)\right)$

$sinx-cos^2x=tan^{\left(2\right)}x-cosx$

$\frac{\left(xy-2x+4y-8\right)}{\left(xy+3x-y-3\right)}\cdot\frac{dy}{dx}=1$

$\left(4b^2+6c\right)^3$