$\frac{x-2}{x-3}>2$
$\left(y^2+4x\right)^3$
$\lim_{x\to-1}\left(x^3-4x+1\right)$
$\left(1+x^{\frac{3}{4}}\right)^2$
$16m^2+16m+3=0$
$\lim_{x\to0}\left(3x^2-6x+1\right)$
$\int x\cos\left(-6x\right)dx$
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