prove : $\frac{b_{1}}{x-a_{1}}+\frac{b_{2}}{x-a_{2}}+...+\frac{b_{k}}{x-a_{k}}=0$ prove that : $\frac{b_{1}}{x-a_{1}}+\frac{b_{2}}{x-a_{2}}+...+\frac{b_{k}}{x-a_{k}}=0$ with $b_{1},b_{2},...,b_{k}>0$ and $a_{1}<a_{2}<...<a_{k}$ has $k -1$ real solutions.
so i think i have managed to prove that though induction and the use of Bolzano's theorem my only problem is that in the end my solution got too big and at some points i think i might have some mistakes.Is there a faster and better way to solve this without induction because i think that bolzano's therorem is needed anyways?(i didnt post my solution since it is too big like 2 sheets of paper.)
 A: Consider $P(x) = (x-a_1)^{ b_1} \cdots (x-a_n)^{ b_n}$
By differentiating $\ln (P(x))$ or otherwise we can show 
$$ H(x) =\dfrac {P' ( x)}{P(x) } = \dfrac{ b_1}{(x-a_1)}+ \cdots \dfrac{b_n}{(x-a_n)}$$
Now $P(x) $ has $n$ distinct roots excluding multiplicities. By Rolle's theorem we have at $ (n-1)$ roots of $P'(x)$ that are different from the roots of $P(x)$. Hence $ H(x) $ has $ (n-1)$ real roots and thus has all its roots real .
A: Multiply with all denomiators. You get a polynomial
  \begin{align*}
    p(x) = \sum_{i=1}^k b_i\prod_{j\neq i} (x-a_j)
  \end{align*}
of degree $k-1$ on the left-hand side.
None of the $a_i$ is a zero of that polynomial because for $x=a_i$ the left-hand side simplifies to $b_i \prod_{j\neq i} (x-a_j)$ and $a_i$ is not a zero of that simplified term.
Therefore, the zeros of the polynomial are the zeros of the original left-hand side.
Because of $a_i < a_{i+1}$ and $b_i>0$ you have
  \begin{align*}
    (-1)^{k-i} p(a_i) > 0.
  \end{align*}
for $k=1,\ldots,k$ and therefore $k$ sign-changes for the continuous function $p(x)$. In between these sign changes there are $k-1$ zeros.
A: Set $f(x)=\frac{b_1}{x-a_1}+\frac{b_2}{x-a_2}+\cdots+\frac{b_k}{x-a_k}$.
One can easily see that the equation has no solutions in $(-\infty, a_1)$ or in $(a_k,+\infty)$ because $f(x)$ is strictly negative on the first interval and strictly positive on the second.
Also, $f(x)$ is strictly decreasing on all intervals where it is defined, so it can have at most one solution in each of the intervals $(a_i, a_{i+1})$ for $i=1,2,\ldots,k-1$. All that is left to be proven is that it actually has one solution in each of those $k-1$ intervals.
So, pick $i\in\{1,2,\ldots,k-1\}$. Note that: $$\lim_{x\to a_i+0}f(x)=+\infty$$ and $$\lim_{x\to a_{i+1}-0}f(x)=-\infty$$
Thus, we can choose $\varepsilon$, $0\lt\varepsilon\lt \frac{a_{i+1}-a_i}{2}$ such that $f(a_i+\varepsilon)\gt 0$ and $f(a_{i+1}-\varepsilon)\lt 0$. Thus, there is a point $\xi\in(a_i+\varepsilon, a_{i+1}-\varepsilon)\subset(a_i, a_{i+1})$ such that $f(\xi)=0$ (by Bolzano's theorem).
A: Hint (following achille hui's comment):
Compare $\lim_{x\to a_j^+}$ and $\lim_{x\to a_{j+1}^-}$ of the expression.
