b, ĐK \(x\in\left\{-4;-5;-6;-7\right\}\)
pt trên tđ với :
\(\left(\dfrac{1}{x+4}-\dfrac{1}{x+5}\right)+\left(\dfrac{1}{x+5}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+7}\right)=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)