ĐKXĐ: \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\ge0\\\sqrt{x+7}=b>0\end{matrix}\right.\) \(\Rightarrow3x-4=\frac{25a^2-b^2}{8}\)
Phương trình trở thành:
\(5a-b=\frac{25a^2-b^2}{8}\Leftrightarrow\left(5a-b\right)\left(5a+b\right)=8\left(5a-b\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5a-b=0\\5a+b=8\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}5\sqrt{x-1}=\sqrt{x+7}\\5\sqrt{x-1}+\sqrt{x+7}=8\end{matrix}\right.\)
TH1: \(5\sqrt{x-1}=\sqrt{x+7}\Leftrightarrow25\left(x-1\right)=x+7\Rightarrow x=\frac{4}{3}\)
TH2: \(5\sqrt{x-1}+\sqrt{x+7}=8\)
\(\Leftrightarrow5\sqrt{x-1}-5+\sqrt{x+7}-3=0\)
\(\Leftrightarrow\frac{5\left(x-2\right)}{\sqrt{x-1}+1}+\frac{x-2}{\sqrt{x-7}+3}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{5}{\sqrt{x-1}+1}+\frac{1}{\sqrt{x-7}+3}\right)=0\)
\(\Rightarrow x=2\)
