Ta có: \(9x^2-1+\left(3x-1\right)\left(x+2\right)=0\)
\(< =>\left(3x-1\right)\left(3x+1\right)+\left(3x-1\right)\left(x+2\right)=0\)
\(< =>\left(3x-1\right)\left(3x+1+x+2\right)=0\)
\(< =>\left(3x-1\right)\left(4x+3\right)=0\)
\(< =>\left\{{}\begin{matrix}3x-1=0\\4x+3=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\frac{1}{3}\\x=-\frac{3}{4}\end{matrix}\right.\)
Vậy tập nghiệm của pt đã cho là: \(S=\left\{\frac{1}{3};-\frac{3}{4}\right\}\)
\(9x^2-1+\left(3x-1\right)\left(x+2\right)=0\Leftrightarrow9x^2-1+3x^2+6x-x-2=0\Leftrightarrow12x^2+5x-3=0\Rightarrow\Delta=5^2-12\times\left(-3\right)=64>0\Rightarrow x1=\frac{1}{3},x2=-\frac{3}{4}\)
\(9x^2-1+\left(3x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)+\left(3x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left[\left(3x+1\right)+\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1+x+2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(4x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\4x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\4x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=\frac{-3}{4}\end{matrix}\right.\)
