$\left(\frac{2}{3}x^2+\frac{1}{2}y^3\right)^4$
$6\left(2\right)^x-73=23$
$-2x^3+7x^2-3x-12$
$3x^2+33x+15=0$
$2\left(x\right)\:\left(5\right)$
$2a+6+3a+4$
$a-b\:=c+d$
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