Question

a) Chứng minh:

 x+2\(\sqrt{2x-4}\)=(\(\sqrt{2}\)+\(\sqrt{x+2}\))\(^2\) với x\(\ge\)2.

b) Rút gọn biểu thức:

 \(\sqrt{x+2\sqrt{2x-4}}\)+\(\sqrt{x-2\sqrt{2x-4}}\) với x\(\ge\)2

Answer 1

a/ Sai đề. 

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

b/ \(M=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{\left(\sqrt{2}+\sqrt{x-2}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{x-2}\right)^2}\)

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

1. Nếu \(2\le x\le4\) thì \(M=\sqrt{2}+\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}=2\sqrt{2}\)

2. Nếu \(x>4\) thì \(M=\sqrt{2}+\sqrt{x-2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\)