Question

Giải phương trình: \(4x^2+\frac{1}{x^2}+7=8x+\frac{4}{x}\)

Answer 1

ĐKXĐ: ...

\(4x^2+\frac{1}{x^2}-4\left(2x+\frac{1}{x}\right)+7=0\)

Đặt \(2x+\frac{1}{x}=a\Rightarrow a^2=4x^2+\frac{1}{x^2}+4\Rightarrow4x^2+\frac{1}{x^2}=a^2-4\)

\(a^2-4-4a+7=0\)

\(\Leftrightarrow a^2-4a+3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x+\frac{1}{x}=1\\2x+\frac{1}{x}=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x^2-x+1=0\\2x^2-3x+1=0\end{matrix}\right.\)