\(\dfrac{x+b}{x-5}+\dfrac{x+5}{x-b}=2\)

\(\Rightarrow\dfrac{x+b-\left(x-5\right)}{x-5}+\dfrac{x+5-\left(x-b\right)}{x-b}=0\)

\(\Rightarrow\dfrac{b+5}{x-5}+\dfrac{b+5}{x-b}=0\)

\(\Rightarrow\left(b+5\right)\left(\dfrac{1}{x-5}+\dfrac{1}{x-b}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}b+5=0\\\dfrac{1}{x-5}+\dfrac{1}{x-b}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=-5\\x-b+x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=-5\\x=\dfrac{b+5}{2}\end{matrix}\right.\)

+) With b = -5 

=> \(\dfrac{x-5}{x-5}+\dfrac{x+5}{x+5}=2\)    (Its true)

So with b = -5 and  x \(\in R\) then equation is satisfy

+) With \(x=\dfrac{b+5}{2}\) 

=> \(\dfrac{\dfrac{b+5}{2}+b}{\dfrac{b+5}{2}-5}+\dfrac{\dfrac{b+5}{2}+5}{\dfrac{b+5}{2}-b}=\dfrac{\dfrac{3b+5}{2}}{\dfrac{b-5}{2}}+\dfrac{\dfrac{b+15}{2}}{\dfrac{-b+5}{2}}=\dfrac{3b+5}{b-5}+\dfrac{b+15}{5-b}\)

\(=\dfrac{3b+5-\left(b+15\right)}{b-5}=\dfrac{2b-10}{b-5}=2\)   (It's true)

So the pairs of numbers satisfying the equation (x;b) are \(\left(\infty;-5\right)\) ;  \(\left(\dfrac{b+5}{2};\infty\right)\)

+) With b = -5 

=> 

    (Its true)

So with b = -5 and  x 

 then equation is satisfy

+) With 

=> 

   (It's true)

So the pairs of numbers satisfying the equation (x;b) are 

 ;