Lời giải:
ĐK: \(x\geq 1\)
Ta có:
PT \(\Leftrightarrow 2x^2+5x-1=7\sqrt{(x-1)(x^2+x+1)}\)
Đặt \(\sqrt{x^2+x+1}=a; \sqrt{x-1}=b(a,b\geq 0)\)
\(\Rightarrow 2a^2+3b^2=2x^2+5x-1\). PT trở thành:
\(2a^2+3b^2=7ab\)
\(\Leftrightarrow 2a^2+3b^2-7ab=0\)
\(\Leftrightarrow (2a-b)(a-3b)=0\) \(\Rightarrow \left[\begin{matrix} 2a=b\\ a=3b\end{matrix}\right.\)
Nếu \(2a=b\Leftrightarrow 2\sqrt{x^2+x+1}=\sqrt{x-1}\)
\(\Rightarrow 4(x^2+x+1)=x-1\)
\(\Rightarrow 4x^2+3x+5=0\)
\(\Rightarrow 3x^2+(x+\frac{3}{2})^2+\frac{11}{4}=0\) (vô lý)
Nếu \(a=3b\Leftrightarrow \sqrt{x^2+x+1}=3\sqrt{x-1}\)
\(\Rightarrow x^2+x+1=9(x-1)\)
\(\Rightarrow x^2-8x+10=0\Rightarrow x=4\pm \sqrt{6}\) (đều thỏa mãn)
Vậy............
