Question

Solve for $x:\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x},[a,b,x\ne 0,x\ne -(a+b\left)\right]$

Answer 1

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

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

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

$\Rightarrow -ab=ax+bx+x^2$

$\Rightarrow x^2+(a+b)x+ab=0$

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

$\Rightarrow x+a=0,x+b=0$

$\therefore x=-a,-b$