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\(A=\left(\dfrac{4\sqrt{xy}+x-2\sqrt{xy}+y}{2\left(x-y\right)}\right)\cdot\dfrac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\left(x-y\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=1\)
đk : \(x\ge0;y\ge0;x\ne y\)
A = \(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}=\dfrac{2\sqrt{xy}}{x-y}\)
\(\Leftrightarrow\) \(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{2\sqrt{xy}}{x-y}\)
\(\Leftrightarrow\) \(\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{2\sqrt{xy}}{x-y}\)
\(\Leftrightarrow\) \(\dfrac{x-\sqrt{xy}-\sqrt{xy}-y}{x-y}=\dfrac{2\sqrt{xy}}{x-y}\)
\(\Rightarrow\) \(x-2\sqrt{xy}-y=2\sqrt{xy}\) \(\Leftrightarrow\) \(x-y=4\sqrt{xy}\)
\(\Leftrightarrow\) A = \(\dfrac{2\sqrt{xy}}{4\sqrt{xy}}=\dfrac{1}{2}\)
không biết sai chỗ nào ??? sao bài làm lại trái với câu hỏi thế này ???
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: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
a) \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)
b) \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)
d) \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)