a: ĐKXĐ: x>=0; x<>4

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

\(=\dfrac{x-2\sqrt{x}+2x+4\sqrt{x}-3x-4}{x-4}\cdot\dfrac{\sqrt{x}}{2}\)

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

b: \(M=P\cdot Q=\dfrac{\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{1-5\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)

\(M\left(M-1\right)=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-5x-x-3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)\left(-6x-2\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}+1\right)^2}\)

\(=\dfrac{\sqrt{x}\left(5\sqrt{x}-1\right)\left(6x+2\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}+1\right)^2}\)

TH1: M>=căn M

=>M^2>=M

=>M^2-M>=0

=>5*căn x-1>=0

=>x>=1/25 và x<>4

TH2: M<căn M

=>5căn x-1<0

=>x<1/25

Kết hợp ĐKXĐ, ta được: 0<=x<1/25

Với \(x\ge0;x\ne4\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{x-4}{\sqrt{x}-2}\)

\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}-\sqrt{x}-2-3\sqrt{x}+2}{x-4}.\dfrac{x-4}{\sqrt{x}-2}\)

\(=\dfrac{2x-4\sqrt{x}}{x-4}.\dfrac{x-4}{\sqrt{x}-2}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=2\sqrt{x}\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-3\sqrt{x}-2}{x-4}\right):\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\\ =\dfrac{x+2\sqrt{x}+x-\sqrt{x}-2\sqrt{x}+2-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\times\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\\ =\dfrac{2x-4\sqrt{x}}{\sqrt{x}-2}\times\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\times\dfrac{1}{\sqrt{x}-2}=\dfrac{2\sqrt{x}}{\sqrt{x}-2}\)

a: \(M=1:\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{3x}{2\left(x-4\right)}+\dfrac{1}{2\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{4-2\sqrt{x}}{1}\)

\(=1:\left(\dfrac{2\sqrt{x}-4-3x+\sqrt{x}+2}{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{-2\left(\sqrt{x}-2\right)}{1}\)

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

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

b: M=20

=>\(-4\left(x-4\right)\left(\sqrt{x}-2\right)=-60x+60\sqrt{x}-40\)

=>\(x\sqrt{x}-2x-4\sqrt{x}+8=-15x+15\sqrt{x}-10\)

=>\(x\sqrt{x}+13x-19\sqrt{x}+18=0\)

=>\(x\in\varnothing\)

\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

a: Ta có: \(P=\left(\dfrac{x-2\sqrt{x}+4}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{x+4}{x-4}\right)\)

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

\(=\dfrac{x-2\sqrt{x}+4}{1}\cdot\dfrac{\sqrt{x}+2}{x+2\sqrt{x}}\)

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

b: \(P-2=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}}>0\forall x\) thỏa mãn ĐKXĐ

nên P>2

Ta có: \(P=\left(\dfrac{\sqrt{x}}{x-4}-\dfrac{2}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)

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

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

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

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

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

\(\left(\dfrac{2}{\sqrt{x^2}-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right):\left(\sqrt{x}-\dfrac{x-4}{\sqrt{x}+2}\right)\) (ĐK: \(x>0;x\ne4\))

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

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

\(=-\dfrac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}:2\)

\(=-\dfrac{1}{\sqrt{x}}:2\)

\(=-\dfrac{1}{\sqrt{x}}\cdot\dfrac{1}{2}\)

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

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.

HOÀNG THỊ LAN HƯƠNG SAI, vì ta có công thức \(\sqrt{A^2}=\left|A\right|\) chứ không phải\(\sqrt{A^2}=A\)đâu. Giả sử \(\sqrt{\left(-2\right)^2}=2\)chứ nó đâu bằng -2 đâu