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

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

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

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

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

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

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

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

\(=1\)

\(DK:x\ge\frac{1}{2}\)

PT

\(\Leftrightarrow\left(4x-4\right)+\left(\sqrt{2x-1}-1\right)=0\)

\(\Leftrightarrow4\left(x-1\right)+\frac{2\left(x-1\right)}{\sqrt{2x-1}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(4+\frac{2}{\sqrt{2x-1}+1}\right)=0\)

Vi \(4+\frac{2}{\sqrt{2x-1}+1}>0\)

\(\Rightarrow x=1\left(n\right)\)

Vay nghiem cua PT la \(x=1\)

Bác nào ko thích liên hợp thì thử cách em nhé!

Đặt ẩn phụ hoàn toàn:

ĐK: \(x\ge\frac{1}{2}\).Đặt \(\sqrt{2x-1}=a\ge0\).

\(PT\Leftrightarrow2\left(2x-1\right)+\sqrt{2x-1}-3=0\)

\(\Leftrightarrow2a^2+a-3=0\Rightarrow\orbr{\begin{cases}a=1\\a=-\frac{3}{2}\left(L\right)\end{cases}}\)

Với a = 1 suy ra x = 1(TM)

Vậy...

Is that true/