ĐK:\(x\le-1;x\ge-\frac{1}{4}\)
Ta có \(\sqrt{4x^2+5x+1}+3=2\sqrt{x^2-x+1}+9x\)
\(\Leftrightarrow\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}=9x-3\)
\(\Leftrightarrow\frac{\left(4x^2+5x+1\right)-\left(4x^2-4x+4\right)}{\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}}=9x-3\)
\(\Leftrightarrow\frac{9x-3}{\sqrt{4x^2+5x+1}+2\sqrt{x^2-x+1}}-9x+3=0\)
\(\Leftrightarrow\left(9x-3\right)\left(\frac{1}{\sqrt{4x^2+5x+1}+2\sqrt{x^2-x+1}}-1\right)=0\)
\(\Leftrightarrow9x-3=0\)(PT còn lại vô nghiệm)
\(\Leftrightarrow x=\frac{1}{3}\)
