Question

A = \(\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\). Rút gọn 

Answer 1

Điều kiện : x > 1/2

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

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

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

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

+) Nếu  \(\sqrt{2x-1}\ge1\) => 2x - 1 > 1 => x > 1 thì \(A\sqrt{2}=\sqrt{2x-1}+1-\sqrt{2x-1}+1=2\)

=> \(A=\sqrt{2}\)

+) Nếu \(\sqrt{2x-1}