$\sqrt{x}+3^2$
$3x^2-3xy+6x^3$
$\frac{6x^3-24}{x-2}$
$3-tan^2\left(\theta\:\:+\frac{\pi\:\:}{4}\right)=6.89$
$\left(y+xe^2\right)dx+\left(x+4y^3\right)dy=0$
$2y+3\le2.y$
$\left(3y+8a-9y+9\right)-\left(8a+9y-10\right)$
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