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

=\(\frac{\sqrt{x-2}-2+\sqrt{x-2}+2}{\frac{2}{x}-1}\)=\(\frac{2\sqrt{x-2}}{\frac{2}{x}-1}\)=\(\frac{-2x}{\sqrt{x-2}}\)

6\(C=\frac{\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+2+4\sqrt{x-2}}}{\sqrt{\frac{4}{x^2}-\frac{4}{x}+1}}\) Điều kiện xác định :\(\hept{\begin{cases}x>2\\x\ne6\end{cases}}\)

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

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

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

-Vì x>2 nên \(\frac{2}{x}< \frac{2}{2}=1\)\(\Rightarrow\frac{2}{x}-1< 0\)

\(\sqrt{x-2}\ge0\)nên\(\sqrt{x-2}+2>0\)

Do đó \(C=\frac{\left|\sqrt{x-2}-2\right|+\sqrt{x-2}+2}{1-\frac{2}{x}}\)

*Với x<6 và x>2 \(\Rightarrow x-2< 4\)\(\Rightarrow\sqrt{x-2}< \sqrt{4}=2\)

\(\Rightarrow\sqrt{x-2}-2< 0\)

Cho nên \(C=\frac{2-\sqrt{x-2}+\sqrt{x-2}+2}{1-\frac{2}{x}}\)

\(=\frac{4}{\frac{x-2}{x}}\)

\(=\frac{4x}{x-2}\)

*Với x>6 (không cho x=6 vì để C xác định) 

\(\Rightarrow\sqrt{x-2}>\sqrt{4}=2\)\(\Rightarrow\sqrt{x-2}-2>0\)

Cho nên \(C=\frac{\sqrt{x-2}-2+\sqrt{x-2}+2}{1-\frac{2}{x}}\)

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

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

Lưu ý là không nên để căn ở mẫu.

ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

Ta có \(A=\left(\frac{1}{\sqrt{x}-1}+\frac{x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}-\frac{x-\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

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

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

a: \(B=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-4}\cdot\dfrac{\sqrt{x}+2}{x+16}=\dfrac{1}{\sqrt{x}-2}\)

b: Khi x=9 thì B=1/(3-2)=1

\(A=\dfrac{\sqrt{x}+2+x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\dfrac{x-1-x+\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+3}\)

\(=\dfrac{x+3}{\sqrt{x}+3}\)

a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

b:Sửa đề: 2A

2A=2căn x+5

=>(2căn x+2)/căn x=2căn x+5

=>2x+5căn x-2căn x-2=0

=>2x+3căn x-2=0

=>(căn x+2)(2căn x-1)=0

=>x=1/4