What is the answer to 1/m+n+x= 1/m+1/n+1/x ?

The answer to 1/m+n+x= 1/m+1/n+1/x is x= 1/n ,x=-n;\quad n\ne 0

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1/m+n+x= 1/m+1/n+1/x

\frac{1}{m} + n + x = \frac{1}{m} + \frac{1}{n} + \frac{1}{x}

x = \frac{1}{n}, x = - n; n \neq = 0

Solution steps

\frac{1}{m} + n + x = \frac{1}{m} + \frac{1}{n} + \frac{1}{x}

Multiply by LCM

n^{2} x + n x^{2} = x + n

Solve

n^{2} x + n x^{2} = x + n : x = \frac{1}{n}, x = - n; n \neq = 0

x = \frac{1}{n}, x = - n; n \neq = 0

a^{- m} = \frac{1}{a ^{m}}

\frac{3 x + 9}{2} = \frac{2}{3 x + 5}

\frac{31}{4 x} = 1300000

\frac{4 x ^{2}}{( x - 3 )} = \frac{4 x ^{2}}{( x - 3 )}

x, y^{2} = \frac{x ^{2}}{3 - x}

What is the answer to 1/m+n+x= 1/m+1/n+1/x ?
The answer to 1/m+n+x= 1/m+1/n+1/x is x= 1/n ,x=-n;\quad n\ne 0