Question

Solve it :-.

1/(a+b+x)=1/a +1/b +1/c ​

Answer 1

GiveN :-

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

To FinD :-

• To simplify and solve out for x .

SolutioN :-

Here a quadratic equation is given to us and we need to solve out and find the value of x. Let's simplify this step by step . Firstly take all the terms that has x on one side that is LHS or RHS .By this we have ,

$\sf:\implies \pink{\dfrac{1}{a+b+x}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x} }\\\\\sf:\implies \dfrac{1}{a+b+x}-\dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b} \\\\\sf:\implies \dfrac{x -(a+b+x)}{x(a+b+x)}=\dfrac{a+b}{ab}\\\\\sf:\implies\dfrac{x -a-b-x}{x(a+b+x)}=\dfrac{a+b}{ab}\\\\\sf:\implies \dfrac{-a-b}{x(a+b+c)}=\dfrac{a+b}{ab}\\\\\sf:\implies \dfrac{-(a+b)}{x(a+b+c)} =\dfrac{a+b}{ab} \\\\\sf:\implies \dfrac{-1}{x^2+ax+bx}=\dfrac{1}{ab}\\\\\sf:\implies x^2+ax+bx=-ab \\\\\sf:\implies x^2+ax+bx+ab=0\\\\\sf:\implies x(x+a)+b(x+a)=0\\\\\sf:\implies (x+a)(x-b)=0 \\\\\sf:\implies\underset{\blue{\sf Required\ Values}}{\underbrace{\boxed{\pink{\frak{ x = (-a) \ , (-b)}}}}}$