Question

tìm x

Answer 1

đk x > = 1 

\(\sqrt{x-1-2\sqrt{x-1}+1}=2\sqrt{x-1}-3\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\sqrt{x-1}-3\Leftrightarrow\left|\sqrt{x-1}-1\right|=2\sqrt{x-1}-3\)

đk \(2\sqrt{x-1}-3\ge0\)

\(\left[{}\begin{matrix}\sqrt{x-1}-1=2\sqrt{x-1}-3\\\sqrt{x-1}-1=3-2\sqrt{x-1}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=2\\3\sqrt{x-1}=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{25}{9}\left(loai\right)\end{matrix}\right.\)