Ta có:

\(x=\frac{2}{2\sqrt[3]{2}+2\sqrt[3]{4}}=\frac{1}{\sqrt[3]{2}+\sqrt[3]{4}}=\)\(\frac{2\sqrt[3]{2}-2+\sqrt[3]{4}}{6}\)

\(y=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}=\frac{6\left(\sqrt[3]{2}+\sqrt[3]{4}\right)}{6}\)

\(\Rightarrow xy^3-x^3y=xy\left(y^2-x^2\right)=y^2-x^2=\frac{36\left(\sqrt[3]{4}+4+2\sqrt[3]{2}\right)}{36}\)\(-\frac{4\sqrt[3]{4}+4+2\sqrt[3]{2}-8\sqrt[3]{2}+8-4\sqrt[3]{4}}{36}\)\(=\frac{36\sqrt[3]{4}+144+72\sqrt[3]{2}-12+6\sqrt[3]{2}}{36}=\frac{36\sqrt[3]{4}+78\sqrt[3]{2}+132}{36}\)\(=\frac{6\sqrt[3]{4}+13\sqrt[3]{2}+22}{6}\)

a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)

\(=\dfrac{1}{6}\sqrt{6}\)

b: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)

\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)

a: \(A=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}=10\)

b: \(B=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}=-2\sqrt{y}\)

c: \(C=\dfrac{\sqrt{3}-1}{\sqrt{6}-\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)

b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)

c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)

\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)

`[\sqrt{27}-\sqrt{15}]/[3-\sqrt{5}]+4/[2+\sqrt{3}]-6/\sqrt{3}`

`=[\sqrt{3}(3-\sqrt{5})]/[3-\sqrt{5}]+[4(2-\sqrt{3})]/[4-3]-[2\sqrt{3}.\sqrt{3}]/\sqrt{3}`

`=\sqrt{3}+8-4\sqrt{3}-2\sqrt{3}`

`=8-5\sqrt{3}`

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`[x-y]/[\sqrt{x}+\sqrt{y}]-[x\sqrt{y}+y\sqrt{x}]/\sqrt{xy}`    `ĐK:  x,y > 0`

`=[(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})]/[\sqrt{x}+\sqrt{y}]-[\sqrt{xy}(\sqrt{x}+\sqrt{y})]/\sqrt{xy}`

`=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}`

`=-2\sqrt{y}`

ta có : \(A=\dfrac{x+y-2\sqrt{xy}}{\sqrt{x}-\sqrt{y}}-\dfrac{x+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)

\(=\sqrt{x}-\sqrt{y}-\sqrt{x}=-\sqrt{y}\)

ta có \(B=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

\(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(3+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{9-3}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{3+\sqrt{3}+3-\sqrt{3}}{6}=\dfrac{6}{6}=1\)

\(x=\sqrt{2}\)

\(y=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

\(y\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)

\(y\sqrt{2}=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)

\(y\sqrt{2}=\sqrt{7}+1-\sqrt{7}+1\)

\(y\sqrt{2}=2\)

\(y=\dfrac{2}{\sqrt{2}}\)

Thay \(x=\sqrt{2},y=\dfrac{2}{\sqrt{2}}\) vào A ta có:

\(A=\dfrac{\sqrt{2}.\dfrac{2}{\sqrt{2}}-1}{\sqrt{2}+\dfrac{2}{\sqrt{2}}}-\dfrac{1-\sqrt{2}.\dfrac{2}{\sqrt{2}}}{2\sqrt{2}-\dfrac{2}{\sqrt{2}}}\)

\(=\dfrac{2-1}{2\sqrt{2}}-\dfrac{1-2}{\sqrt{2}}\)

\(=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{\sqrt{2}}\)

\(=\dfrac{3\sqrt{2}}{4}\)

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