$y'+\frac{x^2}{3y^2}=0,\:y\left(0\right)=1$
$1728-3375$
$10\cdot10\cdot10$
$\left(2\sqrt{x}-x\right)\left(x+5\right)$
$\frac { 3 } { 2 } - x > x$
$\log_{16}\left(400-x^2\right)=2$
$3x^2-x-3=0$
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