\(\Leftrightarrow\dfrac{x+1}{\left(x+1\right)^2-1}+\dfrac{x+6}{\left(x+6\right)^2-1}=\dfrac{x+2}{\left(x+2\right)^2-1}=\dfrac{x+5}{\left(x+5\right)^2-1}\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)^2-x-1+\left(x+6\right)\left(x+1\right)^2-x-6=\left(x+2\right)\left(x+5\right)^2-x-2+\left(x+5\right)\left(x+2\right)^2-x-5\)
=>(x+1)(x+6)^2+(x+6)(x+1)^2=(x+2)(x+5)^2+(x+2)^2(x+5)
=>(x+1)(x+6)(x+6+x+1)=(x+2)(x+5)(x+5+x+2)
=>(2x+7)[x^2+7x+6-x^2-7x-10]=0
=>(2x+7)=0
=>x=-7/2