Question

plz give Ans of this question, I have many more...... ​

Answer 1

$\boxed{\underline{\textsf{Solution :}}}$

$\textsf{Now,}\:\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}$

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

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

$\to \frac{bx-ab-ax+ab}{bx-b^{2}}+\frac{ax-ab-bx+ab}{ax-a^{2}}=0$

$\to \frac{(b-a)x}{bx-b^{2}} + \frac{(a-b)x}{ax-a^{2}}=0$

$\to \frac{(b-a)x}{bx-b^{2}} - \frac{(b-a)x}{ax-a^{2}}=0$

$\to x\bigg(\frac{1}{bx-b^{2}}-\frac{1}{ax-a^{2}}\bigg)=0$

$\to x\bigg(\frac{ax-a^{2}-bx+b^{2}}{(ax-a^{2})(bx-b^{2})}\bigg)=0$

$\to x\{(a-b)x-(a^{2}-b^{2})\}=0$

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

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

$\textsf{Either x = 0 or, x - (a + b) = 0}$

$\textsf{i.e., x = 0 , x = a + b}$

$\textsf{which is the required solution.}$