a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

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

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

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

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

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

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

Với \(x\ge0;x\ne4\)

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

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

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

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

ĐKXĐ: \(x\ge2\)

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

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

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

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

Xét \(x\ge4\Rightarrow\sqrt{x-2}\ge\sqrt{2}\)

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

Xét \(0\le x< 4\Rightarrow\sqrt{x-2}< \sqrt{2}\)

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

\(=\dfrac{2}{\sqrt{x}-2}:\dfrac{3\sqrt{x}+6-\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

=2

\(A=\dfrac{2}{\sqrt{x}-2}:\dfrac{3\sqrt{x}+6-\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\left(x\ge0;x\ne4\right)\\ A=\dfrac{2}{\sqrt{x}-2}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

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

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

đk : x >= 0 ; x khác 4 

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

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

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

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

ĐKXĐ: x khác 4; x ≥ 0

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

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

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

\(A^2=x+\sqrt{x^2-4}+16.\left(x-\sqrt{x^2-4}\right)-2.4.\sqrt{x^2-\left(\sqrt{x^2-4}\right)^2}\)

\(A^2=x+\sqrt{x^2-4}+16x-16.\sqrt{x^2-4}-8.\sqrt{x^2-x^2+4}\)

\(A^2=17x-15.\sqrt{x^2-4}-16\)

mình làm đến đây đc thôi, sorry

Dễ thây \(x\ge2\)

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

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

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

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

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

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