ĐKXĐ : \(x\inℝ\)
Ta có : \(\dfrac{x^2+4x+5}{x^2-x+5}-\dfrac{3x}{x^2-3x+5}=1\)
\(\Leftrightarrow1+\dfrac{5x}{x^2-x+5}-\dfrac{3x}{x^2-3x+5}=1\)
\(\Leftrightarrow x.\left(\dfrac{5}{x^2-x+5}-\dfrac{3}{x^2-3x+5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{5}{x^2-x+5}=\dfrac{3}{x^2-3x+5}\left(1\right)\end{matrix}\right.\)
Phương trình (1) <=> 5(x2 - 3x + 5) = 3(x2 - x + 5)
<=> 2x2 - 12x + 10 = 0
<=> x2 - 6x + 5 = 0
<=> (x - 1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Tập nghiệm \(S=\left\{0;1;5\right\}\)