1. If (x^2–y^2) is direct proportional to (x^2+y^2), show that
a. y is dirctly proportional to x
b. (x–y) is dirctly proportional to (x+y)
2. If (x^3–1÷y^3) is direct proportional to (x^3+1÷y^3), show that
a. y is direct proportional to 1÷x
b.(x–1÷y) is direct proportional to (x+1÷y)
Variation
2010-06-11 2:52 am
回答 (2)
2010-06-11 3:23 am
✔ 最佳答案
As follows:圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/01-107.jpg
圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/02-92.jpg
2010-06-10 19:24:48 補充：
For Q1, k=/= -1
For Q2, k=/= 1
2010-06-10 21:14:55 補充：
http://i707.photobucket.com/albums/ww74/stevieg90/01-107.jpg
http://i707.photobucket.com/albums/ww74/stevieg90/02-92.jpg
2010-06-11 3:10 am
a. y is dirctly proportional to x
(x^2-y^2) = k(x^2 + y^2)
x^2 - kx^2 = y^2 + ky^2
x^2(1-k) = y^2(1+k)
x^2 = (1+k)y^2/(1-k)
x = √(1+k)/(1-k) y
For (1+k)/(1-k) > 0, (1+k)>0 and (1-k)=/=0
2010-06-10 19:50:50 補充：
Sorry for missing the negative case ...
(x^2-y^2) = k(x^2 + y^2)
x^2 - kx^2 = y^2 + ky^2
x^2(1-k) = y^2(1+k)
x^2 = (1+k)y^2/(1-k)
x = √(1+k)/(1-k) y
For (1+k)/(1-k) > 0, (1+k)>0 and (1-k)=/=0
2010-06-10 19:50:50 補充：
Sorry for missing the negative case ...
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