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

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

Tại sao xét  x≥4 vậy bạn.

Vì hai vế đều dương nên bình phương hai vế, ta được:

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

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

        \(=2x+2\sqrt{x^2-4\left(2x-4\right)}=2x+2\sqrt{x^2-8x+16}\)

         =2x + 2√ (x-4)^2 = 2x + 2|x-4|

Đến đây bạn tự làm tiếp nha (với x>2)

Sửa đề: x-4

\(A=\dfrac{x-2\sqrt{x}+x+4\sqrt{x}+4+2x+8}{x-4}=\dfrac{4x+2\sqrt{x}+12}{x-4}\)

a) \(3\sqrt{2x}-4\sqrt{2x}+8-2\sqrt{x}\)

\(=-\left(4\sqrt{2x}-3\sqrt{2x}\right)+8-2\sqrt{x}\)

\(=-\sqrt{2x}-2\sqrt{x}+8\) 

b) \(3\sqrt{2x}-\sqrt{72x}+3\sqrt{18x}+18\)

\(=3\sqrt{2x}-6\sqrt{2x}+3\cdot3\sqrt{2x}+18\)

\(=3\sqrt{2x}-6\sqrt{2x}+9\sqrt{2x}+18\)

\(=\left(3+9-6\right)\sqrt{2x}+18\)

\(=6\sqrt{2x}+18\)

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

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

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

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

* Trường Hợp 1: \(\sqrt{x-2}\ge\sqrt{2}\) => \(H=\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\)

* Trường Hợp 2: \(\sqrt{x-2}< \sqrt{2}\) => \(H=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)

Nguyễn Hoàng Tiến làm thế là gần đúng hết rồi 

trường hợp 2 điều kiện của nó phải là : \(0\le\sqrt{x-2}\le\sqrt{2}\)

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc đề dễ hiểu hơn bạn nhé.

1)\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{26^2}=\sqrt{5}-2+26=24-\sqrt{5}\)

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

3) \(=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)\(=\left[{}\begin{matrix}1\left(x>1\right)\\-1\left(x< 1\right)\end{matrix}\right.\)

4) \(=\sqrt{\left(\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}=\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}=2\sqrt{\dfrac{1}{2}}=\sqrt{2}\)

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

3. \(\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{\sqrt{x^2-2.x.1+1^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{|x-1|}{x-1}=\left[{}\begin{matrix}x-1>0\left(x>1\right)\\x-1< 0\left(x< 1\right)\end{matrix}\right.=\left[{}\begin{matrix}=1\\=\dfrac{x+1}{x-1}\end{matrix}\right.\)