সমাধান সেট নির্ণয় কর: $\frac{1}{x}+\frac{1}{a}+\frac{1}{b}=\frac{1}{x+a+b}$

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

বা, $\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{x+a+b}-\dfrac{1}{x}$

বা, $\dfrac{b+a}{ab}=\dfrac{x-1\left(x+a+b\right)}{x\left(x+a+b\right)}$

বা, $\dfrac{b+a}{ab}=\dfrac{x-x-a-b}{x\left(x+a+b\right)}$

বা, $\dfrac{a+b}{ab}=\dfrac{-a-b}{x\left(x+a+b\right)}$

বা, $\dfrac{\left(a+b\right)}{ab}=\dfrac{-\left(a+b\right)}{x\left(x+a+b\right)}$

বা, $\dfrac{1}{ab}=\dfrac{-1}{x\left(x+a+b\right)}$

বা, $x\left(x+a+b\right)=-ab$

বা, $x^2+ax+bx=-ab$

বা, $x^2+ax+bx+ab=0$

বা, $x\left(x+a\right)+b\left(x+a\right)=0$

বা, $\left(x+a\right)\left(x+b\right)=0$

হয়,
$\left(x+a\right)=0$

$\therefore x=-a$

অথবা,
$\left(x+b\right)=0$

$\therefore x=-b$

সুতরাং নির্ণেয় সমাধান সেট: $\left\lbrace-a,-b\right\rbrace$ [Answer]