$\frac{x^2}{x-3}>x+1$
$3\left(3\left(x\right)^3-5\left(x\right)^2+6\left(x\right)\right)$
$1.5^{\frac{3}{2}}\cdot2$
$\left(36x^4-18x^2-18\right)$
$\frac{y^2}{y-6}$
$72y+43y+24x+32y+21x+12x$
$y^2=\frac{3x\left(x+2\right)}{2}$
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