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simplify [ (3x^(3/2)Y^3)/(x^2y^-1/2) ]^2
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\[[\frac{ 3*x^{3/2} *y^3}{ (x^{2}*y^{-1/2}) }]^{2}\]
sorry its all to the -2
\[[3*x^{\frac{ 3 }{ 2 }-2}*y^{3-\frac{ (-1) }{ 2 }}]^{-2}\]
\[3x \frac{3y^{3} }{2 } {-4} y\]
\[[3*x^{\frac{ -1 }{ 2 }}*y^{\frac{ 7 }{ 2 }}]^{-2}\]
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\[3^{-2}*x^{(\frac{ -1 }{ 2 }*(-2))}*y^{(\frac{ 7 }{ 2 }*(-2))}=\frac{ 1 }{ 9 }*x^{1}*y^{-7}\]
\[\frac{ x }{ 9*y ^{7} }\] @Ssusu
thank uu @gorv
\[3x \frac{ 11 }{ 2 }y2\]
@Ssusu always welcme :)
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