Question

$solve \: \frac{1}{(a + b + c)} = \frac{1}{a} + \frac{1}{b } + \frac{1}{x}$
I wanna know that how I can do that by method attached in notebook.
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Answer 1

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

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☛ -b & -a

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

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

➨ $\dfrac{x - a + b + x}{x(a + b+ x)} = \dfrac{a + b}{ab}$

➨ $\dfrac{- 1( a +b )}{x(a + b + \: x )} = \dfrac{a + b}{ab}$

➨ $\dfrac{ - 1(a + b)}{( a+b )} = \dfrac{x(a + b + x)}{ab}$

➨ $- 1 = \dfrac{x(a + b + x)}{ab}$

➨ $- ab = (a + b)x + {x}^{2}$

➨${x}^{2} + (a + b)x + ab = 0$

(Hence, a = 1 , b = (a+b) , c = ab)

Using, D = B² + 4 AC

➨ (a+b)² - 4 × 1 × ab

⇾(a+b)² - 4ab

⇾ (a-b)²

⇾ $\dfrac{ - b + \sqrt{d} }{2a} \: & \dfrac{ - b - \sqrt{d} }{2a}$

⇾ $\dfrac{ - a - b + a - b}{2 \times 1} \: & \dfrac{ - a - b - a + b}{2 \times 1}$

⇾ $\dfrac{ - 2b}{2}$ & $\dfrac{ - 2a}{2}$

✒ So, it will be, -b & -a .

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Latex error part refers to the attachment