Solve the equation $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x+4}=0$ Solve the equation $$\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0.$$
For $x\ne -4;-3;-2;-1;0$ we have $$(x+1)(x+2)(x+3)(x+4)+x(x+2)(x+3)(x+4)+x(x+1)(x+3)(x+4)+\text{...}=0$$
Most likely that's not the author's intention. I have tried to substitute $t=x+2$ to get $$\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}=0$$ which actually isn't easier to work with than the original problem.

If $t$ is a solution, then so is $-t$, hence it is worth tring to write things in terms of $u:=t^2$. Note that
$$\frac1{t-2}+\frac1{t+2}=\frac{(t+2)+(t-2)}{(t-2)(t+2)}=\frac{2t}{u-4} ,$$
$$\frac1{t-1}+\frac1{t+1}=\frac{(t+1)+(t-1)}{(t-1)(t+1)}=\frac{2t}{u-1} ,$$
and of course
$$\frac1t=\frac{2t}{2u} $$
so that we want to solve
$$\frac1{u-4}+\frac1{u-1}+\frac1{2u}=0 $$
This gives you only a quadratic numerator.
