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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: \(\dfrac{y-2\sqrt{y}+1}{\sqrt{y}-1}=\sqrt{y}-1\)
\(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{x-1-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{x-1}\right)\)
\(=\left(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Ta có: \(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)^2}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right):\left(\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)^2}:\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
d) Ta có: \(D=\left(\sqrt{x}+\dfrac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\dfrac{x}{\sqrt{xy}+y}+\dfrac{y}{\sqrt{xy}-x}-\dfrac{x+y}{\sqrt{xy}}\right)\)
\(=\left(\dfrac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\dfrac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\dfrac{\left(x+y\right)\left(x-y\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\left(\dfrac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-x^2+y^2}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\dfrac{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2-x^2+y^2}{\sqrt{xy}\left(\sqrt{x}-y\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\dfrac{-\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{xy}\cdot\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{-\sqrt{xy}\left(x+y\right)}\)
\(=-1\)
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(\text{a) }\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\\ =\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)\left(x-y\right)}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}-y\sqrt{y}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\\ =\sqrt{xy}\)
\(\text{b) }\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(\text{c) }\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\\ =\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}\\ =\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}\\ =\dfrac{\sqrt{y}-1}{x-1}\)
a)\(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\dfrac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{x}\sqrt{y}+y\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}+y\)
\(=x+\sqrt{xy}+y-x+2\sqrt{xy}+y\)
\(=3\sqrt{xy}+2y\)
\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-1\right)\\ =\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}-1}\right)\\ =\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\\ =\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\\ =\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
a: Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b: Ta có: \(\left(\sqrt{x}+1\right)\cdot A=x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\cdot\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=x\)
\(\Leftrightarrow x-2\sqrt{x}+1=0\)
\(\Leftrightarrow x=1\left(loại\right)\)
a: Ta có: \(A=\left(\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1}{x-1}\)
\(=\dfrac{x+\sqrt{x}+1+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+1}{1}\)
\(=x+2\sqrt{x}+1\)
\(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{x-1}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)