Question

[Nguyễn Linh]

  Đk:?

Answer 1

[HT.Phong (9A5)]

\(\sqrt{x^2-x}=\sqrt{3-x}\) 

ĐK: \(\left\{{}\begin{matrix}x^2-x\ge0\\3-x\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le0\end{matrix}\right.\\x\le3\end{matrix}\right.\) \(\Leftrightarrow x\le3\)  

\(\Leftrightarrow x^2-x=3-x\)

\(\Leftrightarrow x^2-x+x=3\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\left(tm\right)\)

Vậy: ...