Question

\(x^4-5x^3-12x^2-5x+1=0\)

Answer 1

Với \(x=0\) không phải nghiệm

Với \(x\ne0\) chia 2 vế cho \(x^2\) ta được:

\(x^2-5x-12-\dfrac{5}{x}+\dfrac{1}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}+2\right)-5\left(x+\dfrac{1}{x}\right)-14=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-5\left(x+\dfrac{1}{x}\right)-14=0\)

Đặt \(x+\dfrac{1}{x}=t\)

\(\Rightarrow t^2-5t-14=0\Rightarrow\left[{}\begin{matrix}t=7\\t=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=-2\\x+\dfrac{1}{x}=7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=0\\x^2-7x+1=0\end{matrix}\right.\) (bấm máy)