1/ab+1/bc+1/ac
=(c/abc)+(a/abc)+(b/abc)
=(a+b+c)/(abc)
(x+y)/(3xy²)-(x-y)/(2x²y)+1/(6xy)
=2x(x+y)/(6x²y²)-3y(x-y)/(6x²y²)+xy/(6x²y²)
=[(2x²+2xy)-(3xy-3y²)+xy]/(6x²y²)
=(2x²+3y²)/(6x²y²)
1/ab+1/bc+1/ac=c/abc+a/abc+b/abc=(a+b+c)/abc
(a+b+c)/abc
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