Có \(\left(x^2+4x+8\right)^2+3x\cdot\left(x^2+4x+8\right)+2x^2=0\) 0
\(_{\Leftrightarrow\left[\left(x^2+4x+8\right)^2+x\cdot\left(x^2+4x+8\right)\right]+\left[2x\cdot\left(x^2+4x+8\right)+2x^2\right]=0}\)
\(\Leftrightarrow\left(x^2+4x+8\right)\cdot\left(x^2+4x+8+x\right)+2x\cdot\left(x^2+4x+8+x\right)=0\)
\(\Leftrightarrow\left(x^2+5x+8\right)\cdot\left(x^2+4x+8+2x\right)=0\)
\(\Leftrightarrow x^2+5x+8=0\)hoặc \(x^2+6x+8=0\)
+) \(x^2+5x+8=0\Leftrightarrow x^2+2\cdot\frac{5}{2}x+\frac{25}{4}+\frac{7}{4}=0 \)
\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2+\frac{7}{4}=0\)( vô lý vì (x + 5/2)^2 + 7/4 >0 với mọi x) => loại
+) \(x^2+6x+8=0 \Leftrightarrow\left(x^2+2x\right)+\left(4x+8\right)=0\)
\(\Leftrightarrow x\cdot\left(x+2\right)+4\left(x+2\right)=0 \Leftrightarrow\left(x+2\right)\cdot\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}}\)
Vậy ...............
