rút gọn \(C=\left(\frac{2}{\sqrt{xy}}+\frac{1}{x}+\frac{1}{y}\right):\frac{\sqrt{xy}.\left(x+y\right)-xy}{x\sqrt{x}+y\sqrt{y}}\)
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Ta có :
Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)
=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)
=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)
=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)
=\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
=\(4-2\sqrt{4-3}\)
=\(4-2\)
=\(2\)
=>\(A=\sqrt{2}\)
\(P=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\frac{1-xy+x+y+2xy}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}.\)
\(P=\frac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1+x+y+xy}\)
\(P=\frac{2\sqrt{x}}{1+x+y+xy}\)Với ĐK \(x\ge0\) và \(y\ge0\)Và \(xy\ne1\)
Nguyễn Ngọc Anh Minh bạn làm sai rồi kìa bước cuối cùng vẫn còn \(2y\sqrt{x}\)
\(A=\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}-\frac{x+y}{\sqrt{xy}}\right)\)
\(=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\frac{x\left(\sqrt{xy}-x\right)\sqrt{xy}+y\left(\sqrt{xy}+y\right)\sqrt{xy}-\left(x+y\right)\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}{\sqrt{xy}\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x^2y-x^2\sqrt{xy}+xy^2+y^2\sqrt{xy}-y^2\sqrt{xy}+x^2\sqrt{xy}}{xy^2-x^2y}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy^2-x^2y}{xy^2+x^2y}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}{xy\left(x+y\right)}\)
\(=\sqrt{y}-\sqrt{x}\)
Điều kiện xác định: \(xy>0\)
Ta có:
\(C=\left(\frac{2}{\sqrt{xy}}+\frac{1}{x}+\frac{1}{y}\right)\div\frac{\left(x+y\right)\sqrt{xy}-xy}{x\sqrt{x}+y\sqrt{y}}\)
\(C=\frac{2\sqrt{xy}+y+x}{xy}\div\frac{\left(x-\sqrt{xy}+y\right)\sqrt{xy}}{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}\)
\(C=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy}\cdot\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
\(C=\left(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\right)^3\)