Question

Pls answer............​

Answer 1

hope this helps u...

Answer 2

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

$\textsf{Now,}\:\frac{1}{(a+b+x)}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$

$\to \frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{1}{b}$

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

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

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

$\to \frac{-1}{x(a+b+x)}=\frac{1}{ab}$

$\textsf{cancelling (a + b) from both sides}$

$\to x(a+b+x)=-ab$

$\to ax+bx+x^{2}+ab=0$

$\to x^{2}+ax+bx+ab=0$

$\to x(x+a)+b(x+a)=0$

$\to (x+a)(x+b)=0$

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

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

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