Question

Giải phương trình sau:

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

Answer 1

theo minh nghĩ \(=\sqrt{x-2}=\frac{\sqrt{x-1}}{\sqrt{x-1}}\)

con lai ban tư tính nhé =\(-\frac{1}{2}\)

Answer 2

ĐK : \(x\ge1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=\sqrt{x-1}\\1-\sqrt{x-1}=\sqrt{x-1}\end{matrix}\right.\)

\(\Leftrightarrow1=2\sqrt{x-1}\Leftrightarrow1=4\left(x-1\right)\Leftrightarrow x=\frac{5}{4}\left(tm\right)\)