ĐKXĐ: \(x\ne\left\{0;-1;-2;-3;-4;-5;-6;-7\right\}\)
\(\dfrac{x+1}{x\left(x+2\right)}+\dfrac{x+6}{\left(x+5\right)\left(x+7\right)}=\dfrac{x+2}{\left(x+1\right)\left(x+3\right)}+\dfrac{x+5}{\left(x+4\right)\left(x+6\right)}\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{x+2}+\dfrac{1}{x+5}+\dfrac{1}{x+7}=\dfrac{1}{x+1}+\dfrac{1}{x+3}+\dfrac{1}{x+4}+\dfrac{1}{x+6}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+7}-\dfrac{1}{x+6}=\dfrac{1}{x+3}-\dfrac{1}{x+2}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(\Leftrightarrow\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{\left(x+4\right)\left(x+5\right)}-\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{6\left(2x+7\right)}{x\left(x+1\right)\left(x+6\right)\left(x+7\right)}=\dfrac{-2\left(2x+7\right)}{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)}\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\\dfrac{3}{x\left(x+1\right)\left(x+6\right)\left(x+7\right)}=-\dfrac{1}{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)}\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow\dfrac{3}{\left(x^2+7x\right)\left(x^2+7x+6\right)}=-\dfrac{1}{\left(x^2+7x+10\right)\left(x^2+7x+12\right)}\)
Đặt \(x^2+7x=t\)
\(\Rightarrow\dfrac{3}{t\left(t+6\right)}=-\dfrac{1}{\left(t+10\right)\left(t+12\right)}\)
\(\Rightarrow3\left(t+10\right)\left(t+12\right)+t\left(t+6\right)=0\)
\(\Rightarrow t^2+18t+90=0\)
\(\Rightarrow\left(t+9\right)^2+9=0\) (vô nghiệm)
Vậy pt có nghiệm duy nhất \(x=-\dfrac{7}{2}\)
