\(ĐKXĐ:\orbr{\begin{cases}x\ge3+\sqrt{3}\\x\le3-\sqrt{3}\end{cases}}\)
\(\Leftrightarrow\sqrt{\left(3x-7\right)^2}-1=3\sqrt{x^2-6x+6}\)
\(\Leftrightarrow\left|3x-7\right|-1=3\sqrt{x^2-6x+6}\)
- Với \(x\ge3+\sqrt{3}\)
\(\Leftrightarrow3x-8=3\sqrt{x^2-6x+6}\)
\(\Leftrightarrow9x^2-48x+64=9\left(x^2-6x+6\right)\)
\(\Rightarrow x=-\frac{10}{3}\left(l\right)\)
- Với \(x\le3-\sqrt{3}\)
\(\Leftrightarrow2-x=\sqrt{x^2-6x+6}\)
\(\Leftrightarrow x^2-4x+4=x^2-6x+6\)
\(\Rightarrow x=1\) ( t/m)
Chúc bạn học tốt !!!
