ĐKXĐ: \(x\ne\left\{-4;-3;-2;-1\right\}\)
\(\dfrac{x^2+2x+1-x}{x+1}+\dfrac{x^2+3x+2-x}{x+2}-\dfrac{x^2+4x+3-x}{x+3}-\dfrac{x^2+5x+4-x}{x+4}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2-x}{x+1}+\dfrac{\left(x+1\right)\left(x+2\right)-x}{x+2}-\dfrac{\left(x+1\right)\left(x+3\right)-x}{x+3}-\dfrac{\left(x+1\right)\left(x+4\right)-x}{x+4}=0\)
\(\Leftrightarrow\dfrac{-x}{x+1}+\dfrac{-x}{x+2}+\dfrac{x}{x+3}+\dfrac{x}{x+4}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+2\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{1}{x^2+5x+4}=\dfrac{-1}{x^2+5x+6}\)
\(\Rightarrow x^2+5x+6=-x^2-5x-4\)
\(\Rightarrow x^2+x+1=0\) (vô nghiệm)
Vậy pt có nghiệm duy nhất \(x=0\)
