\(x^4-104+13x^3-8x=0\)
\(\Leftrightarrow x^4+13x^3-8x-104=0\)
\(\Leftrightarrow x^3\left(x+13\right)-8\left(x+13\right)=0\)
\(\Leftrightarrow\left(x+13\right)\left(x^3-8\right)=0\)
\(\Leftrightarrow\left(x+13\right)\left(x-2\right)\left(x^2+2x+4\right)=0\)
Ta có : \(x^2+2x+4=\left(x^2+2x+1\right)+3\)
\(=\left(x+1\right)^2+3\ge3\forall x\)
\(\Rightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)
Đúng 0
Bình luận (0)
