<s>[INST] 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.
 [/INST] You can write it as:
$$\sum_{j=0}^4\prod_{\begin{matrix}i=0\\i\ne j\end{matrix}}^4(x+i)=0$$
which might make it easier to type into an equation expander. Otherwise, note that expanded it gives:
$$5x^4+40x^3+105x^2+100x+24=0$$
(I believe anyway I did not check this by hand). This does in fact have four real roots.
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