Question

Tìm các giá trị của x sao cho:

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

Answer 1

ĐK: \(x\ge1\)

Bình phương 2 vế ta có:

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

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

\(\Leftrightarrow x-2\sqrt{x-1}-x+1+2\sqrt{x-1}-1=0\)

\(\Leftrightarrow0=0\)(đúng) 

Vậy x>=1

Answer 2

dk :x>=1

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

xet 2 th => tim x