Question

giải phương trình \(x^2-x-\dfrac{1}{x}+\dfrac{1}{x^2}-10=0\)

Answer 1

\(x^2-x-\dfrac{1}{x}+\dfrac{1}{x^2}-10=0\)

\(\Rightarrow\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)-10=0\)

Đặt: \(x+\dfrac{1}{x}=t\) ta có: \(\left(x+\dfrac{1}{x}\right)^2=t^2\Leftrightarrow x^2+2+\dfrac{1}{x^2}=t^2\Leftrightarrow x^2+\dfrac{1}{x^2}=t^2-2\)

\(\Rightarrow t^2-2-t-10=0\)

\(\Rightarrow t^2-t-12=0\)

\(\Rightarrow t^2-4t+3t-12=0\)

\(\Rightarrow t\left(t-4\right)+3\left(t-4\right)=0\)

\(\Rightarrow\left(t+3\right)\left(t-4\right)=0\Leftrightarrow\left[{}\begin{matrix}t=-3\\t=4\end{matrix}\right.\)

Thay vào rồi giải tiếp nha bạn