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) \(A=2\sqrt{7}+3\sqrt{7}-2\sqrt{7}\\ A=3\sqrt{7}\)

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

a) \(A=2\sqrt{7}+3\sqrt{7}-2\sqrt{7}\\ A=3\sqrt{7}\)

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

\(A=\dfrac{x-2\sqrt{xy}+y+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ A=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}-\sqrt{x}+\sqrt{y}\\ A=\sqrt{x}+\sqrt{y}-\sqrt{x}+\sqrt{y}=2\sqrt{y}\)

Đề sai

\(A=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}+\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)

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

\(=2\sqrt{x}\)

Ta có:

\(VT=\left(\frac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\right)\\ =\left(\frac{\sqrt{x}\cdot\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\right)\\ =\left(\frac{\sqrt{xy}\left[\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2\right]}{\sqrt{xy}}\right)\\ =x-y=VP\left(đpcm\right)\)

Vậy với x>0, y>0 ta có đpcm

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

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

= \(x-y=x-y\)

ĐK:  \(x,y>0;x\ne y\)

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

\(=\frac{\sqrt{x^2y}-\sqrt{xy^2}}{\sqrt{x}-\sqrt{y}}\)

\(=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}=VP\)

\(\Rightarrow\)đpcm

Ta có: \(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}\)

TK nha!

 ta có:\(\frac{\left(x\sqrt{y}+y\sqrt{x}\right)\cdot\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=x-y\)

vậy.....

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

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

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

\(=x-y\)( đpcm )