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.
 A: 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 $-2\pm\sqrt{\frac{15\pm\sqrt{145}}{10}}$.
A: 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. $$5t^2 + 25t + 24=0$$
A: 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.
A: 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.
A: Let $f$ be the conveniently translated function defined by :
$$f(t):=\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}$$
What I would like to underline is that a graphical representation provides a good insight for such questions. In particular, it can help to anticipate (or verify)  a certain number of results, in particular the fact that the roots of $f(t)=0$ are all real, separated by integers $-2,-1,0,1,2$ and symmetric with respect to $0$ due to the fact that $f$ is an odd function.

Indeed, this function has a decreasing behavior (the derivative $f'$ is clearly negative) from $+\infty$ to $+\infty$ on each of the intervals: $[k,k+1], \ k=-2,-1,0,1$ delimitated by vertical asymptotes ; the two external branches do not provide any supplementary root.
A: 
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
$(y-1)y(y+1)(y+2) + (y-2)y(y+1)(y+2) + (y-2)(y-1)(y+1)(y+2) + (y-2)(y-1)y(y+2) + (y-2)(y-1)y(y-1) =0$
$(y^2-1)(y^2+2y) + (y^2-4)(y^2+ y) + (y^2-4)(y^2 - 1) + (y^2 -4)(y^2-y) + (y^2-1)(y^2 - y)=0$
$(y^2-1)[(y^2 + 2y)+(y^2 - 2y)] + (y^2-4)[(y^2 + y)+(y^2-y)] + (y^2 -4)(y^2 -1) = 0$
$(y^2-1)2y^2 + (y^2-4)2y^2 + (y^2 -4)(y^2 -1) = 0$
$2y^2(2y^2 - 5) + (y^4 -5y^2 +4) = 0$
$5y^4 - 15y^2 + 4 = 0$
$y^2 = \frac {15 \pm \sqrt{15^2-80}}{10}= \frac {15\pm \sqrt {145}}{10}$
$y = \pm\frac {15\pm \sqrt {145}}{10}$
$x = \pm \frac {15\pm \sqrt {145}}{10}-2$
