2007-02-02 17:15:13 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(x^-1+y^-1)^-1 =
1/(1/x + 1/y) =
1/((x + y)/xy) =
xy/(x + y)
2007-02-02 17:30:29 · answer #1 · answered by Helmut 7 · 1⤊ 0⤋
First two ways are, of course, equivalent.
Here's another way:
Multiply numerator and denominator by xy, the common denominator. Recall that xy/xy = 1, so it maintains 'equivalency' of the expression.
[1/(1/x+1/y) ]*(xy/xy) = xy/(xy/x+xy/y) = xy/(y+x) ta da!
It would benefit your understanding to see how all three methods are equivalent.
2007-02-02 17:29:07 · answer #2 · answered by modulo_function 7 · 0⤊ 0⤋
(x^-1+y^-1)
= 1/x+1/y
= (x+y)/(xy)
(x^-1+y^-1)^-1
= (xy)/(x+y)
2007-02-02 17:19:20 · answer #3 · answered by sahsjing 7 · 0⤊ 0⤋
1/(1/x+1/y) = 1/[(x+y)/xy] = xy/(x+y)
2007-02-02 17:23:35 · answer #4 · answered by Div 2 · 0⤊ 0⤋
1/(1/x+1/y)
1/x+1/y=(y+x)/xy
So 1/(1/x+1/y)=xy/(y+x)
2007-02-02 17:19:20 · answer #5 · answered by Professor Maddie 4 · 0⤊ 0⤋