\(x^2-7x+\sqrt{x^2-7x+8}=12\)
\(x^2-7x-12+\sqrt{x^2-7x+8}=0\)
\(x^2-7x+8-20+\sqrt{x^2-7x+8}=0\)
Đặt : \(\sqrt{x^2-7x+8}=t\left(đk:t>0\right)\)
\(\Rightarrow x^2-7x+8=t^2\)
\(\Rightarrow\)Phương trình trở thành : \(t^2+t-20=0\)
\(\Rightarrow\orbr{\begin{cases}t=4\left(tm\right)\\t=-5\left(L\right)\end{cases}}\)
Với \(t=4\Rightarrow\sqrt{x^2-7x+8}=4\)
\(\Rightarrow x^2-7x+8=16\)
\(\Rightarrow x^2-7x+8-16=0\)
\(\Rightarrow x^2-7x-8=0\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-1\end{cases}}\)
\(x^2-7x+\sqrt{x^2-7x+8}=12\)
\(\Leftrightarrow\sqrt{x^2-7x+8}=12-x^2+7x\)
\(\Leftrightarrow\sqrt{x^2-7x+8}-4=8-x^2+7x\)
\(\Leftrightarrow\frac{x^2-7x+8-16}{\sqrt{x^2-7x+8}+4}=-\left(x-8\right)\left(x+1\right)\)
\(\Leftrightarrow\frac{\left(x-8\right)\left(x+1\right)}{\sqrt{x^2-7x+8}+4}+\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+1\right)\left(\frac{1}{\sqrt{x^2-7x+8}+4}+1\right)=0\)
Dễ thấy: \(\frac{1}{\sqrt{x^2-7x+8}+4}+1>0\)
\(\Rightarrow\orbr{\begin{cases}x-8=0\\x+1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=8\\x=-1\end{cases}}\)
