\(\dfrac{1}{x^2+9x+20}-\dfrac{1}{x^2+7x+12}=\dfrac{x^2-2x-33}{x^2+8x+15}\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}-\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{x^2-2x-33}{\left(x+3\right)\left(x+5\right)}\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)}-\dfrac{1}{x+5}-\dfrac{1}{x+3}+\dfrac{1}{x+4}=\dfrac{x^2-2x-33}{\left(x+3\right)\left(x+5\right)}\)
\(\Leftrightarrow\dfrac{x^2-2x-33+2x+8}{\left(x+3\right)\left(x+5\right)}=\dfrac{2}{x+4}\)
\(\Leftrightarrow\dfrac{x^2-25}{\left(x+3\right)\left(x+5\right)}=\dfrac{2}{x+4}\)
\(\Leftrightarrow\dfrac{x-5}{x+3}=\dfrac{2}{x+4}\)
\(\Leftrightarrow x^2-x-20-2x-6=0\)
\(\Leftrightarrow x^2-3x-26=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-26\right)=9+104=113>0\)
Do đó: Phương trình có hai nghiệm phân biệt là
\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{113}}{2}\\x_2=\dfrac{3+\sqrt{113}}{2}\end{matrix}\right.\)
