Bunuel wrote:
If xy>0, which of the following must be positive?
A. \((\sqrt{xy})^2x^3y^2\)
B. \((\sqrt[3]{x})(\sqrt[5]{y})x^3y^2\)
C. \((\sqrt{xy})x^7y^9\)
D. \((\sqrt[3]{x2y})x^2y^4\)
E. \((\sqrt{x3y5})x^2y^5\)
xy > 0
So either x>0, y>0 OR x<0, y<0
A. \((\sqrt{xy})^2x^3y^2\) = \(x^4 * y^3\) so -ve value is possible if y<0, x<0
B. \((\sqrt[3]{x})(\sqrt[5]{y})x^3y^2\) = \((\sqrt[3]{x})(\sqrt[5]{y})(xy)^2x\)
For this option if y<0, x<0, \((\sqrt[3]{x})\) is +ve , \((\sqrt[5]{y})\) is -ve \(x^2y^2\) is -ve & x is -ve. So overall value is -ve.
C. \((\sqrt{xy})x^7y^9\) = \((\sqrt{xy})(xy)^7y^2\).. All the terms are +ve. So overall +ve.
D. \((\sqrt[3]{x2y})x^2y^4\) = \((\sqrt[3]{xy*y})x^2y^4\) . So if y <0 & x<0, \((\sqrt[3]{xy*y})\) is -ve and hence overall -ve is possible.
E. \((\sqrt{x3y5})x^2y^5\) = \((\sqrt{(xy)^3*y^2})(xy)^2y^3\) (+ve) * (+ve) *(-ve) is y<0 & x<0
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