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標題:

X^2-Y^2=k(X^2+Y^2)

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發問:

X^2-Y^2=k(X^2+Y^2) 證明(a) y=kx (b) x-y=k(x+y) 更新: 唔好意思!!! 題目係話 x^2-y^2隨(x^2+y^2)正變

最佳解答:

X2 - Y2 = k(X+ Y)2 (X+Y)(X-Y) = k(X+ Y)2 Suppose X is not equal to -Y (so X+Y is non-zero), we can divide (X+Y) from both side to get (X-Y) = k(X+Y) 2007-04-22 18:06:54 補充： 我和閣下意見一樣認為題目有錯我認為原題目應該是(X+Y)^2 2007-04-23 09:45:12 補充： 正確答案X^2-Y^2=k(X^2+Y^2)(1-k)X^2=(1+k)Y^2Y^2=(1-k)/(1+k)X^2Y= [(1-k)/(1+k)]^(1/2)X or Y= -[(1-k)/(1+k)]^(1/2)Xi.e. Y is directly proportional to X 2007-04-23 09:47:46 補充： part BSince Y is directly proportional to XY = k'X for a constant k'X - Y = (1-k')XX + Y = (1+k')Xwe can express X - Y as [ (1-k')/(1+k') ](X + Y)let k'' = (1-k')/(1+k')X - Y = k'(X + Y)i.e. X - Y is directly proportional to X + Y

其他解答:

樓主呢題係邊本練習書揾架，好似出錯題： X^2-Y^2=k(X^2+Y^2) (1-k)X^2=(1+k)Y^2 Y^2=(1-k)/(1+k)X^2 Y= X((1-k)/(1+k))^0.5 or Y= - X((1-k)/(1+k))^0.5 題目要証y=kx 即係要証 + k= ((1-k)/(1+k))^0.5 - 咁即係要証: k^2=1-k^2 k^2=1/2 k=+ (1/2)^0.5 _ 所以呢題數要有一些特定value既k, y=kx先會成立 但係因為題目冇givenｋ既值 所以係証唔到 2007-04-22 11:16:50 補充： 上便嗰位人兄做乜改咗人地條題目ｘ＾２＋ｙ＾２冇端端點會等於（ｘ＋ｙ）＾２