Question

\(\text{Giải phương trình}:\frac{x^4-x^2+1}{x^3+3x^2-x}=\frac{1}{2}\)

Answer 1

ĐKXĐ: \(x\ne\left\{0;\frac{-3\pm\sqrt{13}}{2}\right\}\)

Phương trình tương đương: \(\frac{x^2+\frac{1}{x^2}-1}{x-\frac{1}{x}+3}=\frac{1}{2}\)

Đặt \(x-\frac{1}{x}=a\Rightarrow x^2+\frac{1}{x^2}=a^2+2\)

Pt trở thành: \(\frac{a^2+1}{a+3}=\frac{1}{2}\)

\(\Leftrightarrow2a^2+2=a+3\)

\(\Leftrightarrow2a^2-a-1=0\)

\(\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{x}=1\\x-\frac{1}{x}=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x-1=0\\2x^2+x-2=0\end{matrix}\right.\) (casio)