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.

Let $y=x+2$. Then your equation becomes$$\frac1{y-2}+\frac1{y-1}+\frac1y+\frac1{y+1}+\frac1{y+2}=0$$or$$\frac{5 y^4-15 y^2+4}{(y-2) (y-1) y (y+1) (y+2)}=0,$$which can be solved using the quadratic formula. Its roots are $\pm\sqrt{\frac{15\pm\sqrt{145}}{10}}$, and therefore the roots of the original equation are $-2\pm\sqrt{\frac{15\pm\sqrt{145}}{10}}$.
