Prove that
Question:

$\frac{a}{(x-b)}+\frac{b}{(x-a)}=2,(x \neq b, a)$

 

Solution:

$\frac{a}{(x-b)}+\frac{b}{(x-a)}=2$

$\Rightarrow\left[\frac{a}{(x-b)}-1\right]+\left[\frac{b}{(x-a)}-1\right]=0$

$\Rightarrow \frac{a-(x-b)}{x-b}+\frac{b-(x-a)}{x-a}=0$

$\Rightarrow \frac{a-x+b}{x-b}+\frac{a-x+b}{x-a}=0$

$\Rightarrow(a-x+b)\left[\frac{1}{(x-b)}+\frac{1}{(x-a)}\right]=0$

$\Rightarrow(a-x+b)\left[\frac{(x-a)+(x-b)}{(x-b)(x-a)}\right]=0$

$\Rightarrow(a-x+b)\left[\frac{2 x-(a+b)}{(x-b)(x-a)}\right]=0$

$\Rightarrow(a-x+b)[2 x-(a+b)]=0$

$\Rightarrow a-x+b=0$ or $2 x-(a+b)=0$

$\Rightarrow x=a+b$ or $x=\frac{a+b}{2}$

Hence, the roots of the equation are $(a+b)$ and $\left(\frac{a+b}{2}\right)$.

 

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