<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] Perhaps this will help: $${1\over x}+{1\over x+4} = {2x+4\over x^2+4x}$$
$${1\over x+1}+{1\over x+3} = {2x+4\over x^2+4x+3}$$
$${1\over x+2} = {x+2\over x^2+4x+4}$$
and now we have $$ {x+2\over x^2+4x+4}+   {2x+4\over x^2+4x} + {2x+4\over x^2+4x+3}=0$$
Clearly $x\ne -2$ and let $t=x^2+4x$ then you have:
$${1\over t+4}+{2\over t+3} + {2\over t}=0$$ i.e. $$5t^2 + 25t + 24=0$$
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