\(\left(x+4\right)\left(3x-1\right)+\left(x^2+8x+16\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+4\right)\left(3x-1\right)=0\\x^2+8x+16=0\end{cases}}\)
Xét PT 1 : \(\left(x+4\right)\left(3x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{1}{3}\end{cases}}\)
Xét PT 2 : \(x^2+8x+16=0\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)
\(\left(x+4\right)\left(3x-1\right)+\left(x^2+8x+16\right)=0\)
\(3x^2-x+12x-4+x^2+8x+16=0\)
\(4x^2+19x+12=0\)
\(x^2+\frac{19}{4}x+3=0\)
\(x^2+4x+\frac{3}{4}x+3=0\)
\(\Rightarrow x\left(x+4\right)+3\left(\frac{1}{4}x+1\right)=0\)
Đến đây mk gợi ý thôi mk đi nghủ đây
