Question

Simplify the equation: $\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x};a+b\ne 0$

Answer 1

$\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$ $a+b\ne 0$

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

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

$\Rightarrow abx=(a+b{)}^{2}x+(a+b)ab+x^2(a+b)+abx$

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

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

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

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

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

$x=-b$ $x=-a$

$x=-a,-b$