\(\Leftrightarrow\left(x^4+4x^3+4x^2\right)=11x^2-8x-4\\ \)
\(\Leftrightarrow\left(x^2+2x\right)^2=11x^2-8x-4\)
\(\Leftrightarrow\left(x^2+2x\right)^2+4\left(x^2+2x\right)+4=\left(4x^2+8x+4\right)+11x^2-8x-4\\ \)
\(\Leftrightarrow\left(x^2+2x+2\right)^2=15x^2=\left(\sqrt{15}x\right)^2\)
\(\Leftrightarrow\left[\begin{matrix}x^2+2x+2=-\sqrt{15}x\left(1\right)\\x^2+2x+2=\sqrt{15}x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2+\left(2+\sqrt{15}\right)x+2=0\Leftrightarrow x^2+\left(2+\sqrt{15}\right)x+\left(\frac{2+\sqrt{15}}{2}\right)^2=\frac{\left(2+\sqrt{15}\right)^2}{4}-2=\frac{11+4\sqrt{15}}{4}\left(3\right)\)\(\left(3\right)\Leftrightarrow\left(x+\frac{2+\sqrt{15}}{2}\right)^2=\frac{11+\sqrt{15}}{4}=\left(\frac{\sqrt{11+\sqrt{15}}}{2}\right)^2\)(4)
\(\left(4\right)\Leftrightarrow\left(x+a\right)^2=b^2\Leftrightarrow\left[\begin{matrix}x+a=-b\\x+a=b\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=\frac{-a-b}{2}\\x=\frac{-a+b}{2}\end{matrix}\right.\) Thay a, b vào rồi rút gọn x đến tối giản:
(2) \(11-4\sqrt{15}\) < 11-4.3<0 => (2) vô nghiệm
