Giải pt: x4+2x3=4x+4

\(x^4+2x^3=4x+4\) \(\Leftrightarrow x^4+2x^3-4x-4=0\) \(\Leftrightarrow\left(x^4+2x^3+2x^2\right)-\left(2x^2+4x+4\right)=0\) \(\Leftrightarrow x^2\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)=0\) \(\Leftrightarrow\left(x^2-2\right)\left(x^2+2x+2\right)=0\) \(\Leftrightarrow x^2-2=0\) ( Vì : \(x^2+2x+2>0\) ) \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)Câu trả lời: \(x^4+2x^3=4x+4\) \(\Leftrightarrow x^4+2x^3-4x-4=0\) \(\Leftrightarrow\left(x^4+2x^3+2x^2\right)-\left(2x^2+4x+4\right)=0\) \(\Leftrightarrow x^2\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)=0\) \(\Leftrightarrow\left(x^2-2\right)\left(x^2+2x+2\right)=0\) \(\Leftrightarrow x^2-2=0\) ( Vì : \(x^2+2x+2>0\) ) \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)