Question

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

Answer 1

Given,

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

$abx=bx(x+a+b)+ax(x+a+b)+ab(x+a+b)$

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

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

$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(a^2+b^2+2ab)\pm \sqrt{{(a^2+b^2+2ab)}^2-4(a+b)(a{b}^2+a^{2}b)}}{2(a+b)}$

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

$\therefore x=-b,-a$