Giải:
\(\left(x^2+\dfrac{1}{x^2}\right)-3\left(x+\dfrac{1}{x}\right)+4=0\left(1\right)\)
ĐKXĐ: \(xe0\)
Đặt \(x+\dfrac{1}{x}=a\)
\(\Leftrightarrow a^2=\left(x+\dfrac{1}{x}\right)^2\)
\(\Leftrightarrow a^2=x^2+2+\dfrac{1}{x^2}\)
\(\Leftrightarrow a^2-2=x^2+\dfrac{1}{x^2}\)
Ta có phương trình:
\(\left(1\right)\Leftrightarrow a^2-2-3a+4=0\)
\(\Leftrightarrow a^2-3a+2=0\)
\(\Leftrightarrow a^2-2a-a+2=0\)
\(\Leftrightarrow a\left(a-2\right)-\left(a-2\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-1=0\\a-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}-1=0\\x+\dfrac{1}{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}-1e0\\x=1\left(TM\right)\end{matrix}\right.\)
Vậy ...
