Question

giải phương trình  \(x^4+x^3-10x^2+x+1=0\)

Answer 1

\(x^4+x^3-10x^2+x+1=0\Leftrightarrow x^2+x-10+\frac{1}{x}+\frac{1}{x^2}=0\Leftrightarrow\left(x+\frac{1}{x}\right)^2+\left(x+\frac{1}{x}\right)-12=0\)(1)

Đặt \(t=x+\frac{1}{x}\), khi đó:

(1) \(t^2+t-12=0\Leftrightarrow\orbr{\begin{cases}t=3\\t=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{x}=3\\x+\frac{1}{x}=-4\end{cases}}}\)

\(\Leftrightarrow\)\(x=\frac{3+\sqrt{5}}{2}\)hoặc \(x=\frac{3-\sqrt{5}}{2}\)hoặc \(x=-2+\sqrt{3}\)hoặc \(x=-2-\sqrt{3}\).