$\left(s+\frac{2}{3}t\right)^2$
$\frac{8x^3+27y^3}{3x}$
$\frac{3}{2}x+4<=10$
$( x ^ { 4 } y ) ( \frac { 3 x ^ { 4 } y ^ { - 3 } z } { 6 x ^ { 8 } y ^ { 2 } z ^ { 4 } } )$
$-\frac{9}{8}n.\frac{4}{3}m^2$
$x^2-12x-121=0$
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