If `(xy)/(x+y)=a`, `(xz)/(x+z)=b` and `(yz)/(y+z)=c`, where `a`, `b` and `c` are other than zero, then `x` equal
a.    `(abc)/(ab+bc+ca)`
b.    `(2abc)/(ab+bc+ca)`
c.    `(2abc)/(ab+ac-bc)`
d.    `(2abc)/(ac+bc-ab)`

Explanation

Here,`(xy)/(x+y) = a`

`=>1/(1/x+1/y) = a`

`=>1/x+1/y = 1/a->(1)`

Similarly, `1/x+1/z = 1/b->(2)`

and `1/y+1/z = 1/c->(3)`

Equation `(1)-`Equation`(3)`,

`1/x-1/z = 1/a-1/c->(4)`

Equation`(2) + `Equation `(4)`

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