\(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{x-2-2\sqrt{2}.\sqrt{x-2}+2}=\text{| }\sqrt{x-2}+\sqrt{2}\text{| }+\text{| }\sqrt{x-2}-\sqrt{2}\text{| }\) +) \(A=\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\left(\text{ }\sqrt{x-2}\text{≥}\sqrt{2}\right)\)

+) \(A=\sqrt{x-2}+\sqrt{2}+\sqrt{2}-\sqrt{x-2}=2\sqrt{2}\left(\sqrt{x-2}< \sqrt{2}\right)\)

Ta có \(A\sqrt{2}=\sqrt{2x+4\sqrt{2x-4}}+\sqrt{2x-4\sqrt{2x-4}}=\sqrt{2x-4+4\sqrt{2x-4}+4}\)

+\(\sqrt{2x-4-4\sqrt{2x-4}+4}=\sqrt{2x-4}+2+\sqrt{2x-4}-2=2\sqrt{2x-4}\)

=> A=\(2\sqrt{x-2}\)

A = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)

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

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

\(\sqrt{2}A=\sqrt{2x+4\sqrt{2x-4}}+\sqrt{2x-4\sqrt{2x-4}}\)

\(\sqrt{2}A=\sqrt{\left(2x-4\right)+2.2\sqrt{2x-4}+4}+\sqrt{\left(2x-4\right)-2.2\sqrt{2x-4}+4}\)

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

\(\sqrt{2}A=|\sqrt{2x-4}+2|+|\sqrt{2x-4}+2|\)

\(\sqrt{2}A=\sqrt{2x-4}+2+|\sqrt{2x-4}-2|\)

Xét 2 trường hợp:

+)\(\sqrt{2x-4}\ge2\)

\(\sqrt{2}A=\sqrt{2x-4}+2+\sqrt{2x-4}-2\)

\(\sqrt{2}A=2\sqrt{2x-4}\)

\(A=\sqrt{2}\sqrt{2x-4}=\sqrt{4x-8}\)

+)\(\sqrt{2x-4}< 2\)

\(\sqrt{2}A=\sqrt{2x-4}+2+2-\sqrt{2x-4}=4\)

Vậy...

( Bạn có thể bình phương lên cũng đc)

Suy ra A=2\(\sqrt{2}\) 

P=\(1+2\sqrt{x}\).

Q=x-1.

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

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

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

Ta có: \(\frac{2x+2}{\sqrt{x}}+\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x^2+\sqrt{x}}{x\sqrt{x}+x}\)

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

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

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

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

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

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

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

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

ĐK:\(x>0\)

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

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

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

\(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)