ĐKXĐ: \(x\ne-1;-2;-3;-4\)
\(\Leftrightarrow\dfrac{4}{x+4}-\dfrac{3}{x+3}=\dfrac{2}{x+2}-\dfrac{1}{x+1}\)
\(\Leftrightarrow\dfrac{4\left(x+3\right)-3\left(x+4\right)}{\left(x+3\right)\left(x+4\right)}=\dfrac{2\left(x+1\right)-\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x}{x^2+7x+12}-\dfrac{x}{x^2+3x+2}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{x^2+7x+12}-\dfrac{1}{x^2+3x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{x^2+7x+12}-\dfrac{1}{x^2+3x+2}=0\left(1\right)\end{matrix}\right.\)
Xét \(\left(1\right)\Leftrightarrow\dfrac{1}{x^2+7x+12}=\dfrac{1}{x^2+3x+2}\Leftrightarrow x^2+7x+12=x^2+3x+2\)
\(\Leftrightarrow4x=-10\Rightarrow x=\dfrac{-5}{2}\)
Vậy pt có 2 nghiệm: \(\left[{}\begin{matrix}x=0\\x=\dfrac{-5}{2}\end{matrix}\right.\)
