$y=\frac{1}{9}\left(4x^2+3\right)^{\frac{3}{2}}$
$y=\left(10x+2\right)^x$
$\left(tan\:alpha\:+\:tan\:beta\right)$
$7x^2+10=5x$
$\left(\frac{1}{2}u-3v\right)^2$
$16y^2+60a+36$
$( 5 x - 1 ) ( 7 x - 5 )$
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