maths .urgent.......

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2013-02-11 13:35:21 補充：
Similarly, (1/x² - 1/y²) ÷ (1/x + 1/y)
= (y²/x²y² - x²/x²y²) ÷ (y/xy + x/xy)
= (y² - x²)/x²y² ÷ (y + x)/xy
= (y + x)(y - x)/x²y² * xy/(y + x)
= (y - x)/xy

2013-02-11 13:35:31 補充：
Similarly, (1/x² - 1/y²) ÷ (1/x - 1/y)
= (y²/x²y² - x²/x²y²) ÷ (y/xy - x/xy)
= (y² - x²)/x²y² ÷ (y - x)/xy
= (y + x)(y - x)/x²y² * xy/(y - x)
= (x + y)/xy

Put x = 1 and y = 2 :

(1/x² + 1/y²) ÷ (1/x + 1/y)
= (1/1² + 1/2²) ÷ (1/1 + 1/2)
= (5/4) ÷ (3/2)
= (5/4) x (2/3)
= 5/6

1/x² - 1/xy + 1/y²
= 1/1² - 1/(1*2) + 1/2²
= 1 - 1/2 + 1/4
= 3/4

Hence, (1/x² + 1/y²) ÷ (1/x + 1/y) ≠ 1/x² - 1/xy + 1/y²

No,the question is (1/x^2+1/y^2)/(1/x+1/y).it is HKCEE1991 mc question 5.

Do you mean 1/x^2 - 1/y^2 ?