Đk x >= 1
\(\sqrt{2x-1}-2\sqrt{x-1}=-1\Leftrightarrow\sqrt{2x-1}=2\sqrt{x-1}-1\)
Bình phương 2 vế ta có :
2x - 1 = 4( x - 1) - 4 \(\sqrt{x-1}\) + 1
=> 2x - 1 = 4x - 4 + 1 - 4 căn( x - 1)
=> 2x - 1 = 4x - 3 - 4 căn ( x - 1)
=> 4 căn ( x - 1) = 2x - 2
=> 2 can ( x - 1) = x - 1
=> 4 ( x - 1) = x^2 - 2x + 1
=> 4x - 4 - x^2 + 2x - 1 = 0
=> -x^2 + 6x - 5 = 0
=> x^2 - 6x + 5 = 0
=> x = 1 hoặc x = 5