Very Basic Math question?

Question

How can I prove that $$\frac{r}{(1-x)^2} + \frac{rx}{x(1-x)} = \frac{r}{x(1-x)^2}$$ I have tried to prove that , but I could not , can someone help me please ?

Thanks

Answer 1

Multiply by $x(1-x)^2$ (assuming $x\notin\{0,1\}$), we get

$$rx+rx(1-x) = r.$$

Or, $$rx(1+1-x) = r.$$

Thus, unless $x(2-x) = 1$ or $r=0$, the above is not true.

So $x=1$ or $r=0$ are only two possible cases when equation may hold.

At $x=1$, LHS and RHS of original equation are unbounded, so only possible case is $r=0$.

Answer 2

Take any reals $r,x$ such that $x \notin \{0,1\}$.

Then $\frac{r}{(1-x)^2} = \frac{rx}{x(1-x)^2}$.

And $\frac{rx}{x(1-x)} = \frac{rx(1-x)}{x(1-x)^2}$.

The sum of these is not $\frac{r}{x(1-x)^2}$.

However, the following is true:

$\frac{r}{(1-x)^2} + \frac{r}{x(1-x)} = \frac{r}{x(1-x)^2}$.

Answer 3

If your original expression to prove is actually: $$\frac{r}{(1-x)^2} + \frac{r}{x(1-x)} = \frac{r}{x(1-x)^2}$$ then it works out as follows: \begin{align} \frac{r}{(1-x)^2} + \frac{r}{x(1-x)} &=\frac{rx}{x(1-x)^2} + \frac{r(1-x)}{x(1-x)^2}\\[1ex] &=\frac{rx +r -rx}{x(1-x)^2}\\[1ex] &= \frac{r}{x(1-x)^2} \end{align}

As observed, though, your original expression doesn't generally work.