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=-n,x=-m;\quad m+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 = - n, x = - m; m + n \neq = 0

Solution steps

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

Multiply by LCM

mn x = n x (x + m + n) + m x (x + m + n) + mn (x + m + n)

Solve

mn x = n x (x + m + n) + m x (x + m + n) + mn (x + m + n) : x = - n, x = - m; m + n \neq = 0

x = - n, x = - m; m + n \neq = 0

x, \frac{2}{x} + \frac{3}{y} = 13

\frac{( x + \frac{1}{x} ) ^{2} - 3}{2 ( x - \frac{1}{x} )} = 4

\frac{x ^{4} + 1}{2 x ^{2}} = \frac{17}{8}

\frac{x ^{2} + 1}{2 x} = 2

\frac{x}{6} - \frac{3}{x} = \frac{1}{2}

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=-n,x=-m;\quad m+n\ne 0