Solve for x :
1/a+b+x = 1/a +1/b +1/x (a≠0, b≠0, x≠0).

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Solution

1/(a+b+x)=1/a + 1/b + 1/x

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

abx=abx+a2x+a2b+b2x+abx+ab2+bx2+ax2+abx
ax2+bx2+a2x+abx+abx+b2x+a2b+ab2=0

x2(a+b)+ax(a+b)+bx(a+b)+ab(a+b)=0

(a+b)(x2+ax+bx+ab)=0

So, x2+ax+bx+ab=0

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

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

x=-a,-b

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Similar questions

x : \frac{1}{(a + b + x)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}

{a \neq 0, b \neq 0, x \neq 0, x \neq - (a + b)}

x : \frac{1}{(a a + b + x)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}, [a \neq 0, b \neq 0, x \neq 0, x \neq - (a + b)]

\frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}; (x \neq 0)

\frac{1}{a} - b + x = \frac{1}{x} + \frac{1}{a} - \frac{1}{b}

a \neq= 0

b \neq 0

x \neq 0

x = - a, x = b)

x : \frac{1}{(a + b + x)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}, [a \neq 0, b \neq 0, x \neq 0, x \neq - (a + b)]

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