Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\left(\dfrac{x^2-3x+3}{x}\right)\left(\dfrac{x^2-2x+3}{x}\right)=2\)
\(\Leftrightarrow\left(x+\dfrac{3}{x}-3\right)\left(x+\dfrac{3}{x}-2\right)-2=0\)
Đặt \(x+\dfrac{3}{x}-3=t\)
\(\Rightarrow t\left(t+1\right)-2=0\Leftrightarrow t^2+t-2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+\dfrac{3}{x}-3=1\\x^2+\dfrac{3}{x}-3=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2-x+3=0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
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