Question

Rút gọn biểu thức: A= \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)

Answer 1

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

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

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

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

\(A^2=2x+2|x-4|\)

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

Answer 2

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

ĐKXĐ :x\(\ge2\)

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

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

* Nếu \(\sqrt{2}-\sqrt{x-2}\ge0\Leftrightarrow x\le4\) thì:

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

* Nếu \(\sqrt{2}-\sqrt{x-2}< 0\Leftrightarrow x>4\) thì :

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