\(\frac{x+\sqrt{5}}{x^2+2\sqrt{5}x+5}=\frac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\frac{1}{x+\sqrt{5}}\)

\(B=\frac{x+\sqrt{5}}{x^2+2\sqrt{5}x+5}=\frac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\frac{1}{x+\sqrt{5}}=\frac{x-\sqrt{5}}{x^2-5}\)

A=căn[(x-5)²]/x-5=|x-5|/x-5

Nếu x>=5 thì A=1

Nếu x<5 thì A=-1

mình trình bày trong hình, hơi mờ bạn thông cảm!

như này là xịn lém rrr. Thănkiu nhìu nhó Nguyen Thu Hong

Sửa đề; \(D=\left(\dfrac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}-2\sqrt{y}}-\dfrac{2\sqrt{xy}}{x-y}\right)\cdot\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

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

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

\(\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{x+\sqrt{x}}-\frac{2}{1-x}\right)\) (ĐKXĐ : \(x>0;x\ne1;x\ne\frac{1}{9}\) )

\(=\left[\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}}{\sqrt{x}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{3\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}\)

\(=\frac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}{3\sqrt{x}-1}\)