zogfly3的部落格

標題:

數學知識交流 － 求值 (2)

發問:

若 x + y = 3，x?+ y? = 24，已知 x, y 為實數，求 (1) x3+ y3 (2) x2- y2 (3) x?- y? 所有可能值之和。

最佳解答:

x + y = 3 ==> x2 + y2 = 9 - 2xy ......(1) x?+ y?= (x2 + y2)2 - 2x2y2 = 24 ......(2) 代 (1) 入 (2) :(9 - 2xy)2 - 2x2y2 = 24 2(xy)2 - 36xy + 57 = 0 xy = 9 ± √210 /2 x(3 - x) = 9 ± √210 /2 x2 - 3x + 9 ± √210 /2 = 0 因 x 為實數 , 只有 x2 - 3x + 9 - √210 /2 = 0 , 故 xy = 9 - √210 /2 , (x - y)2 = x2 + y2 - 2xy = 9 - 4xy ..... (由(1)) = 9 - 4(9 - √210 /2) = 2√210 - 27x - y = ±√ (2√210 - 27) *************************************************************************************** 1)x3 + y3 = (x + y) (x2 - xy + y2) = (3) (9 - 3xy) ..... (由(1)) = 27 - 9 (9 - √210 /2) = 9√210 /2 - 54 所有可能值之和 = 9√210 /2 - 54 2)x2 - y2 = (x - y) (x + y) = ±√ (2√210 - 27) (3)所有可能值之和 = 0 3)x?- y? = (x - y) ( x?+ x3y + x2y2 + xy3 + y?) = (x - y) ( x?+ y?+ xy(x2 + xy + y2) ) = (x - y) ( x?+ y?+ xy ((x + y)2 - xy) ) = ±√ (2√210 - 27) (24 + (9 - √210 /2) (32 - (9 - √210 /2)) ) 所有可能值之和 = 0 註 : x = (1/2) (3 - √ (2√210 - 27) ) 或 x = (1/2) (3 + √ (2√210 - 27) ) y = (1/2) (3 + √ (2√210 - 27) ) 或 y = (1/2) (3 - √ (2√210 - 27) ) 其實 2) 和 3) 題不用計也知是 0 。

其他解答:

• 我要兩條小六的趣味數學題。
• 有點困難的數學問題@1@
• concentration problem

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