Lời giải:
ĐKXĐ: \(x\geq 9\)
Ta có:
\(x+6\sqrt{x-9}=(x-9)+6\sqrt{x-9}+9=(\sqrt{x-9}+3)^2\)
\(\Rightarrow \sqrt{x+6\sqrt{x-9}}=\sqrt{x-9}+3\)
\(x-6\sqrt{x-9}=(x-9)-6\sqrt{x-9}+9=(\sqrt{x-9}-3)^2\)
\(\Rightarrow x-6\sqrt{x-9}=|\sqrt{x-9}-3|\)
Do đó pt tương đương với:
\(6(\sqrt{x-9}+3)+6|\sqrt{x-9}-3|=x+23\)
+) Nếu \(x\geq 18\Rightarrow |\sqrt{x-9}-3|=\sqrt{x-9}-3\)
Khi đó:
\(6(\sqrt{x-9}+3)+6(\sqrt{x-9}-3)=x+23\)
\(\Leftrightarrow 12\sqrt{x-9}=x+23\)
\(\Rightarrow 144(x-9)=(x+23)^2\)
\(\Leftrightarrow x^2-98x+1825=0\)
\(\Rightarrow \left[\begin{matrix} x=25\\ x=73\end{matrix}\right.\) (đều thỏa mãn)
+) Nếu $9\leq x< 18$ thì \(|\sqrt{x-9}-3|=3-\sqrt{x-9}\)
PT trở thành:
\(6(\sqrt{x-9}+3)+6(3-\sqrt{x-9})=x+23\)
\(\Leftrightarrow 36=x+23\Rightarrow x=13\) (thỏa mãn)
Vậy.......
