\(\dfrac{4x}{x^2-4x+7}=\dfrac{3x}{x^2-5x+7}\)
\(\Leftrightarrow4x\left(x^2-5x+7\right)=3x\left(x^2-4x+7\right)\)
\(\Leftrightarrow4\left(x^2-5x+7\right)=3\left(x^2-4x+7\right)\)
\(\Leftrightarrow4x^2-20x+28=3x^2-12x+21\)
\(\Leftrightarrow x^2-8x+7=0\)
\(\Leftrightarrow x^2-7x-x+7=0\)
\(\Leftrightarrow x\left(x-7\right)-\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
Vậy x = 1 hoặc x = 7
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