Question

solve 1/a+b+x=1/a+1/b+1/c

Answer 1

1/a+b+c- 1\x= 1/a+1/b
x-(a+b+x)/x(a+b+x) = b+a/ab
x-a-b-x/ x(a+b+x)= b+a/ab
-a-b/x(a+b+x)= b+a/ab
-1/x(a+b+x)= 1/ab
-ab=x(a+b+x)
-ab= ax+ bx+x^2
x^2+bx+ax+ab
x(x+b) a(x+b)
(x+a)(x+b)
x=-a ...x=-b

Answer 2

Answer:

Given :

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

Now take 1 / x in L.H.S. side :

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

Taking L.C.M and solving it further :

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

$\displaystyle{\dfrac{-1}{(a+b+x)(x)} = \dfrac{1}{ab} }\\\\\\\displaystyle{\dfrac{-ab}{(a+b+x)(x)} = 1}\\\\\\\displaystyle{(a+b+x)(x)+ab=0}$

x² + a x + b x + a b = 0

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

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

x + a = 0  or x + b = 0

x = - a or x = - b .

Therefore , the value of x is - a or - b .