Question

1/a+b+x=1/a+1/b+1/x, x≠0, -(a+b)​

Answer 1

Answer:

x = - a, - b

Step-by-step explanation:

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

Answer 2

GIVEN:-

$\bf \: •Equation:- \large\frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}$

TO FIND OUT:-

$\bf\:• Roots \: of \: the \: given \: equation(x) = {?}$

SOLUTION:-

$\bf \implies \large \frac{1}{a + b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b}$

$\bf \implies \: \large \frac{x - (a + b + x)}{x(a + b + x)} = \frac{b + a}{ab}$

$\bf \implies \: \large \frac{ - (a + b)}{x(a + b + x)} = \frac{a + b}{ab}$

$\bf \implies \: \large \frac{ - 1}{x(a + b + x)} = \frac{1}{ab}$

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

$\bf \implies \: x {}^{2} + (a + b)x + ab = 0$

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

$\bf \implies \: x + a = 0 \: or \: x + b = 0$

$\bf\implies \: x = - a \: or \: x = - b$

$\bf \: Hence \: the \: roots \: of \: the \: given \: equation \: are \: - a \: and - b$