Question

[phước]

\(\frac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}}\)

rút gọn biểu thức

Answer 1

[Tuyển Trần Thị]

\(\frac{A}{\sqrt{2}}\)=\(\frac{\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}}\) (DK \(x\ge1\)

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

vs  \(x\ge2\) \(\frac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{2x-1}+1-\sqrt{2x-1}+1}=\frac{2\sqrt{x-1}}{2}=\sqrt{x-1}\) \(\Rightarrow A=\sqrt{2x-2}\)

vs \(1\le x< 2\) \(\frac{\sqrt{x-1}+1+1-\sqrt{x-1}}{\sqrt{2x-1}+1-1+\sqrt{2x-1}}=\frac{1}{\sqrt{2x-1}}\) \(\Rightarrow A=\frac{\sqrt{2}}{\sqrt{2x-1}}\)

Answer 2

[Tuyển Trần Thị]

\(\sqrt{2X-1}\ge1\Leftrightarrow X\ge1\)NEN SUY RA THEO CACH LAM CUA TO 

THOI U AM BUSY SEE YOU AGAIN