Đặt \(\left\{{}\begin{matrix}x^2-5x+1=a\\x^2-4=b\end{matrix}\right.\) \(\Rightarrow x-1=\dfrac{b-a}{5}\)
Pt trở thành:
\(ab=6\left(\dfrac{b-a}{5}\right)^2\)
\(\Leftrightarrow6a^2-37ab+6b^2=0\)
\(\Leftrightarrow\left(a-6b\right)\left(6a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=6b\\6a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+1=6\left(x^2-4\right)\\6\left(x^2-5x+1\right)=x^2-4\end{matrix}\right.\)
\(\Leftrightarrow...\)
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