<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] 
Most likely that's not the author's intention

It could very well be.  If we replace $y = x + 2$ we have
$\frac 1{y-2} + \frac 1{y-1} +\frac 1y + \frac 1{y+1}+\frac 1{y+2} = 0$ so
$(y-1)y(y+1)(y+2) + (y-2)y(y+1)(y+2) + (y-2)(y-1)(y+1)(y+2) + (y-2)(y-1)y(y+2) + (y-2)(y-1)y(y-1) =0$
$(y^2-1)(y^2+2y) + (y^2-4)(y^2+ y) + (y^2-4)(y^2 - 1) + (y^2 -4)(y^2-y) + (y^2-1)(y^2 - y)=0$
$(y^2-1)[(y^2 + 2y)+(y^2 - 2y)] + (y^2-4)[(y^2 + y)+(y^2-y)] + (y^2 -4)(y^2 -1) = 0$
$(y^2-1)2y^2 + (y^2-4)2y^2 + (y^2 -4)(y^2 -1) = 0$
$2y^2(2y^2 - 5) + (y^4 -5y^2 +4) = 0$
$5y^4 - 15y^2 + 4 = 0$
$y^2 = \frac {15 \pm \sqrt{15^2-80}}{10}= \frac {15\pm \sqrt {145}}{10}$
$y = \pm\frac {15\pm \sqrt {145}}{10}$
$x = \pm \frac {15\pm \sqrt {145}}{10}-2$
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