Question

solve it .......................​

Answer 1

Question :

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

Solve it .

Solution :

$\sf \dfrac{1}{x}+\dfrac{1}{a}+\dfrac{1}{b} = \dfrac{1}{x + a + b} \\ \\ \implies \sf \: \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{x + a + b} - \dfrac{1}{x} \\ \\ \implies \sf \: \dfrac{b + a}{ab} = \dfrac{x - (x + a + b)}{x(x + a + b)} \\ \\ \implies \sf \: \dfrac{b + a}{ab} = \dfrac{x - x - a - b}{ {x}^{2} + ax + bx} \\ \\ \implies \sf \: \dfrac{b + a}{ab} = \dfrac{ - b - a}{ {x}^{2} + ax + bx} \\ \\ \implies \sf \: \dfrac{b + a}{ab} = \dfrac{ - (b + a)}{ {x}^{2} + ax + bx} \\ \\ \implies \sf \: \dfrac{1}{ab} = \dfrac{ - 1}{ {x}^{2} + ax + bx} \\ \\ \implies \sf \: {x}^{2} + ax + bx = - ab \\ \\ \implies \sf \: {x}^{2} + ax + bx + ab = 0 \\ \\ \implies \sf \: x(x + a) + b(x + a) = 0 \\ \\ \implies \sf \: (x + a)(x + b) = 0$

Either ,

x + a = 0

→ x = -a

Or,

x + b = 0

→ x = -b

Therefore,

• x = -a
• x = -b

Answer 2

QuestioN :

$\tt \dagger \: \: \: \: \: \dfrac{1}{x} + \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{x + a + b}$

SolutioN :

We have, Quadratic Equation.

$\tt : \implies \dfrac{1}{x} + \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{x + a + b}$

$\tt : \implies \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{x + a + b} - \dfrac{1}{x}$

$\tt : \implies \dfrac{a + b}{ab} = \dfrac{x - (x + a + b)}{x(x + a + b)}$

$\tt : \implies \dfrac{a + b}{ab} = \dfrac{ \cancel{x} \cancel{ - x } - a - b}{x(x + a + b)}$

$\tt : \implies \dfrac{a + b}{ab} = \dfrac{- a - b}{x(x + a + b)}$

$\tt : \implies \dfrac{a + b}{ab} = \dfrac{- ( a + b)}{x(x + a + b)}$

$\tt : \implies \dfrac{ \cancel{a + b}}{ab} = \dfrac{- \cancel{( a + b)}}{x(x + a + b)}$

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

$\tt : \implies - ab = x(x + a + b)$

$\tt : \implies - ab = {x}^{2} + ax + bx$

$\tt : \implies 0 = {x}^{2} + ax + bx + ab$

$\tt : \implies 0 = x(x + a) + b(x + a)$

$\tt : \implies 0 = (x + a)(x + b)$

$\rule{90}2$

# Either,

→ x + a = 0.

→ x = - a.

# Or,

→ x + b = 0.

→ x = - b.

Therefore, the value of x = - a , - b.