\(ĐKXĐ:x\ge-\frac{2}{3}\)
Ta có : \(4x^2+6x+1=4\sqrt{6x+4}\)
\(\Leftrightarrow4x^2+6x+1+6x+4+4=6x+4+4\sqrt{6x+4}+4\)
\(\Leftrightarrow4x^2+12x+9=\left(\sqrt{6x+4}\right)^2+2.\sqrt{6x+4}.2+2^2\)
\(\Leftrightarrow\left(2x+3\right)^2=\left(\sqrt{6x+4}+2\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=\sqrt{6x+4}+2\left(1\right)\\2x+3=-\sqrt{6x+4}-2\left(2\right)\end{cases}}\)
+) Pt (1) \(\Leftrightarrow\sqrt{6x+4}=2x+1\)
\(\Leftrightarrow\hept{\begin{cases}5x+4=4x^2+4x+1\\x\ge-\frac{1}{2}\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)\left(4x+3\right)=0\\x\ge-\frac{1}{2}\end{cases}}\) \(\Leftrightarrow x=1\) ( Thỏa mãn )
+) Pt (2) \(\Leftrightarrow\sqrt{6x+4}=-2x-5\)
\(\Leftrightarrow\hept{\begin{cases}6x+4=\left(-2x-5\right)^2\\x\le-\frac{5}{2}\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}6x+4=4x^2+25+20x\\x\le-\frac{5}{2}\end{cases}}\) ( Vô nghiệm )
Vậy phương trình đã cho có nghiệm duy nhất \(x=1\)
