\(\left(3x+2\right)^2+2x\left(1-2x\right)+4x\left(x-\dfrac{1}{2}\right)=9\)
\(\Leftrightarrow9x^2+12x+4+2x-4x^2+4x^2-2x-9=0\)
\(\Leftrightarrow9x^2+12x+4-9=0\)
\(\Leftrightarrow\left(3x+2\right)^2-3^2=0\)
\(\Leftrightarrow\left(3x+2-3\right)\left(3x+2+3\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{3}\) hoặc \(x=-\dfrac{5}{3}\)
= (3x)2 + 12x + 4 + 2x - 4x2 + 4x2 - 2x = 9
=> (3x)2 + 12x + 4 = 9
=> (3x + 2)2 = 9
=> 3x + 2 = \(\pm\) 3
+) 3x + 2 = 3 => x = \(\dfrac{1}{3}\)
+) 3x + 2 = -3 => x = \(\dfrac{-5}{3}\)