標題:

數學測驗勁難題.......help下me...........

發問:

比個詳盡過程我....唔該 a(2次方) + b(2次方) + c(2次方) + 42 < ab + 9b + 8c help help 我.....唔該哂............ 更新: 什麼也看不懂......我肯定是42... 更新 2: sor.....唔記得左打個問........係求a、b、c的值..........sor哇....

最佳解答:

consider (a-b/2)^2 = a^2-ab+b^2/4 -----(1) 3(b/2-3)^2 = 3b^2/4-9b+27 ----(2) (c-4)^2 =c^2 -8c +16 -----(3) (1)+(2)+(3), we have (a-b/2)^2+3(b/2-3)^2+(c-4)^2=a^2+b^2+c^2+43 -ab -9b -8c ab+9b+8c = a^2+b^2+c^2 +43 -(red part>=0) ab+9b+8c>a^2+b^2+c^2+43 are you sure the integer is 42 not 43, please reply me (to be continued) 2007-11-07 19:18:48 補充: since 43 42, thenab 9b 8c>a^2 b^2 c^2 43 a^2 b^2 c^2 42 2007-11-07 19:20:23 補充: since 43>42, thenab+9b+8c>a^2+b^2+c^2+43>a^2+b^2+c^2+42 2007-11-07 19:21:21 補充: hope I can help you! 2007-11-07 22:16:21 補充: sorry I just made a mistake in hurryab+9b+8c= a^2+b^2+c^2+43-(red part>=0)thereforeab+9b+8c<=a^2+b^2+c^2+43 2007-11-07 22:16:34 補充: your inequality is not true, you can just by putting a,b andc equal to 1, then you will find you inequality cannot hold. Therefore, I believe you make mistakes(also including writing 42 instead of 43). 2007-11-12 00:09:05 補充: 因為ab+9b+8c = a^2+b^2+c^2 +43 -(a-b/2)^2+3(b/2-3)^2+(c-4)^2所以,若要ab+9b+8c>a^2+b^2+42的話那麼red part就必須等於0所以c-4=0, b/2-3=0, a-b/2=0所以 c=4, b=6, a=3

其他解答:

(a-b/2)^2 = a^2-ab+b^2/4 -----(1) 3(b/2-3)^2 = 3b^2/4-9b+27 ----(2) (c-4)^2 =c^2 -8c +16 -----(3) (1)+(2)+(3), we have (a-b/2)^2+3(b/2-3)^2+(c-4)^2=a^2+b^2+c^2+43 -ab -9b -8c ab+9b+8c = a^2+b^2+c^2 +43 -(red part>=0) ab+9b+8c>a^2+b^2+c^2+43 are you sure the integer is 42 not 43, please reply me

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