Question

1/a+b+x=1/a+1/b+1/x find the value of x

Answer 1

Here's your answer!!

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$= > \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}$


$= > ( \frac{1}{a + b + x} - \frac{1}{x}) = ( \frac{1}{a} + \frac{1}{b} )$


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


$x \: and \: - x \: get \: cancelled$


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


$- (a + b)(ab) = (a + b)( {x}^{2} + ax + bx)$


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


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$\frac{ - (a + b)}{(a + b)} \: get \: cancelled$
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$= > - ab = {x}^{2} + ax + bx$


Now,


$= > {x}^{2} + ax + bx + ab = 0$

$= > x(x + a) + b(x + a)$


$= > (x + a)(x + b)$


Therefore,

$= > x + a = 0 \\ = > x = - a$

And,

$= > x + b = 0 \\ = > x = - b$

Hence,

x can be either ( -a ) or ( -b)

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Hope it helps you!! :)

Answer 2

1/a+b+x = 1/a+1/b+1/x

1/a+b+x = bx+ax+ab/abx

(a+b+x) (bx+ax+ab) = abx

abx+a²x+a²b+b²x+abx+ab²+bx²+ax²+abx = abx

(a+b)x² + (a²+b²)x + 3abx-abx + ab(a+b) = 0

(a+b)x² + (a²+b²)x + 2abx + ab(a+b) = 0

(a+b)x² + (a²+b²+2ab)x + ab(a+b) = 0

(a+b)x² + (a+b)²x + ab(a+b) = 0

(a+b) [x²+(a+b)x+ab] = 0

(a+b) [x²+ax+bx+ab] = 0

(a+b) [x(x+a)+b(x+a)] = 0

(a+b) [(x+a)(x+b) ] = 0

(x+a)(x+b) = 0/(a+b)

(x+a)(x+b)=0

x = -a and x = -b