Question

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

Answer 1

$\frac{ab+bx+ax}{abx}=\frac{1}{a+b+x}$

$a^{2}b+a{b}^2+3abx+b^{2}x+a^{2}x+x^{2}a+x^{2}b=abx$

$x^2(a+b)+x(a+b{)}^2+ab(a+b)=0$

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

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

or, $x=-a,-b$