Question

Rut gon bieu thuc A=\(\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\)

Answer 1

A²=x+\(\sqrt{2x-1}\)+x-\(\sqrt{2x-1}\)- 2\(\sqrt{\left(x+\sqrt{2x-1}\right)\left(x-\sqrt{2x-1}\right)}\)

A²=2x-2\(\sqrt{x^2-2x+1}\)

A²=2x-2(x-1)=1

=>A=1(vì a>0)

Answer 2

Ta có: \(A=\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\)   \(\left(ĐK:x\ge\frac{1}{2}\right)\)

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

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

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

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

   \(\Leftrightarrow A\sqrt{2}=2\)

   \(\Leftrightarrow A=\sqrt{2}\)