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\(N=\dfrac{xy\left(x^{\dfrac{1}{3}}+y^{\dfrac{1}{3}}\right)}{x^{\dfrac{1}{3}}+y^{\dfrac{1}{3}}}=xy\)

\(=\dfrac{xy\left(x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}\right)}{x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}}=xy\)

23 tháng 8 2023

\(A=\dfrac{x^{\dfrac{3}{2}}y+xy^{\dfrac{3}{2}}}{\sqrt{x}+\sqrt{y}}=\left(x+y\right).\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\).

3 tháng 8 2020

Ta có : \(P=\frac{\frac{\left(x-y\right)^3}{\left(\sqrt{x}+\sqrt{y}\right)^3}+2x\sqrt{x}+y\sqrt{y}}{x\sqrt{x}+y\sqrt{y}}+\frac{3\left(\sqrt{xy}-y\right)}{x-y}\)

=> \(P=\frac{\frac{\left(\sqrt{x}+\sqrt{y}\right)^3\left(\sqrt{x}-\sqrt{y}\right)^3}{\left(\sqrt{x}+\sqrt{y}\right)^3}+2x\sqrt{x}+y\sqrt{y}}{\sqrt{x}^3+\sqrt{y}^3}+\frac{3\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

=> \(P=\frac{\left(\sqrt{x}-\sqrt{y}\right)^3+2x\sqrt{x}+y\sqrt{y}}{\sqrt{x}^3+\sqrt{y}^3}+\frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=> \(P=\frac{x\sqrt{x}-3x\sqrt{y}+3y\sqrt{x}-y\sqrt{y}+2x\sqrt{x}+y\sqrt{y}}{\left(x+y\right)\left(x-\sqrt{xy}+y\right)}+\frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=> \(P=\frac{3x\sqrt{x}-3x\sqrt{y}+3y\sqrt{x}}{\left(x+y\right)\left(x-\sqrt{xy}+y\right)}+\frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=> \(P=\frac{3\sqrt{x}\left(x-\sqrt{xy}+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}+\frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=> \(P=\frac{3\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=> \(P=\frac{3\sqrt{x}+3\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\frac{3\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=3\)

4 tháng 9 2016

Sai đề

a: \(A=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{2}}+y^{\dfrac{1}{3}}\cdot x^{\dfrac{1}{2}}}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}\left(x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}\right)}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}=\left(xy\right)^{\dfrac{1}{3}}\)

b: \(B=\dfrac{x^{3+\sqrt{3}}}{y^2}\cdot\dfrac{x^{-\sqrt{3}-1}}{y^{-2}}=\dfrac{x^{3+\sqrt{3}-\sqrt{3}-1}}{y^{2-2}}=x^2\)