\(A=\left(5-\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\right)\)

\(A=\left[5-\frac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\right]\left[5+\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\right]\)

\(A=\left[5-\sqrt{xy}\right]\left[5+\sqrt{xy}\right]\)

\(A=25-xy\)

vậy \(A=25-xy\)

\(A=\left(5-\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\right)\)

\(A=\left(5-\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\right)\)

\(A=\left(5-\sqrt{xy}\right)\left(5+\sqrt{xy}\right)\)

\(A=25-xy\)

\(A=\left(5-\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\right)\)

\(=\left(5-\frac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\right)\)

\(=\left(5-\sqrt{xy}\right)\left(5+\sqrt{xy}\right)\)

\(=25-xy\)

\(A=\dfrac{x^{\dfrac{5}{4}}y+xy^{\dfrac{5}{4}}}{\sqrt[4]{x}+\sqrt[4]{y}}\\ =\dfrac{xy\left(x^{\dfrac{1}{4}}+y^{\dfrac{1}{4}}\right)}{x^{\dfrac{1}{4}}+y^{\dfrac{1}{4}}}\\ =xy\)

\(B=\left(\sqrt[7]{\dfrac{x}{y}\sqrt[5]{\dfrac{y}{x}}}\right)^{\dfrac{35}{4}}\\= \left(\sqrt[7]{\dfrac{x}{y}\cdot\left(\dfrac{x}{y}\right)^{-\dfrac{1}{5}}}\right)^{\dfrac{35}{4}}\\ =\left(\sqrt[7]{\left(\dfrac{x}{y}\right)^{\dfrac{4}{5}}}\right)^{\dfrac{35}{4}}\\ =\left[\left(\dfrac{x}{y}\right)^{\dfrac{4}{35}}\right]^{\dfrac{35}{4}}\\ =\left(\dfrac{x}{y}\right)^{\dfrac{4}{35}\cdot\dfrac{35}{4}}\\ =\left(\dfrac{x}{y}\right)^1\\ =\dfrac{x}{y}\)

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

\(B=\dfrac{\sqrt{x}-\sqrt{y}}{x\sqrt{x}-y\sqrt{y}}=\dfrac{1}{x+\sqrt{xy}+y}\)

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

\(D=\dfrac{x+\sqrt{5x}+5}{x\sqrt{x}-5\sqrt{5}}=\dfrac{1}{\sqrt{x}-\sqrt{5}}\)

Ta có \(A=\left(\frac{2\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\frac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}\)

         \(=\left(\frac{4\sqrt{xy}+\left(\sqrt{x}-\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)                 (Quy đồng biểu thức đầu và đổi dấu số hạng cuối)

         \(=\left(\frac{4\sqrt{xy}+x-2\sqrt{xy}+y}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

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

          \(=\frac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=1.\)

Vậy giá trị biểu thức \(A=1.\)

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

bài này dài lắm mk ko tiện làm

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

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

a: \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5}+1-\sqrt{5}+1\)

=2

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

d: \(\dfrac{y-2\sqrt{y}+1}{\sqrt{y}-1}=\sqrt{y}-1\)

a) \(\frac{\sqrt{2x^3}}{\sqrt{8x}}=\sqrt{\frac{2x^3}{8x}}=\frac{1}{2}x\)

b) \(\left(3-\sqrt{5}\right)\left(x+\sqrt{5}\right)=3^2-\left(\sqrt{5}\right)^2=9-5=4\)

c) \(\sqrt{\frac{3x^2y^4}{27}}=0\)

\(y\ne0\)

Thì \(\sqrt{\frac{3x^2y^4}{27}}=\frac{1}{3}xy^2\)

e) \(\frac{y}{x^2}\sqrt{\frac{36x^4}{y^2}}=\frac{y}{x^2}.\frac{6x^2}{\left|y\right|}=\frac{6y}{\left|y\right|}\)

Vì y < 0 nên \(\left|y\right|=-y\)

Vậy \(\frac{6y}{\left|y\right|}=\frac{6y}{-y}=-6\)

f) \(\frac{\sqrt{99999999}}{\sqrt{11111111}}=\sqrt{\frac{99999999}{11111111}}=\sqrt{9}=3\)

\(=\sqrt{x\sqrt{x^{1+\dfrac{1}{2}}}}:x^{\dfrac{5}{8}}\)

\(=\sqrt{x\cdot x^{\dfrac{1}{2}\cdot\dfrac{3}{2}}}:x^{\dfrac{5}{8}}\)

\(=\sqrt{x^{1+\dfrac{3}{4}}}:x^{\dfrac{5}{8}}\)

\(=x^{\dfrac{1}{2}\cdot\dfrac{7}{4}}:x^{\dfrac{5}{8}}=x^{\dfrac{7}{8}-\dfrac{5}{8}}=x^{\dfrac{1}{4}}=\sqrt[4]{x}\)

=>A