Re: [解題] 高一數學求整數解

看板tutor作者wettland5566 (韋特蘭5566)時間15年前 (2009/07/19 06:48)推噓5(5推 0噓 12→)

留言17則, 6人參與討論串2/3 (看更多)

※ 引述《alistair (在乎妳感受)》之銘言： : 1.年級：高一數學 : 2.科目：數學 : 3.章節：第一章 : 4.題目： : a. 求 X+Y=X^2-X*Y+Y^2的整數解(X,Y) : b. x,y滿足 x^3-6*x^2+13*x-2006=0 : y^3+3*y^2+4*y+1998=0 : 求(x+y)=? : 謝謝!! 第一題 x+y = x^2-xy+y^2 x^2-xy+(1/4)*y^2 + (3/4)*y^2 -x-y = 0 (x-y/2)^2 - (x-y/2) + 1/4 + (3/4)*y^2 - (3/2)*y + 3/4 = 1/4 + 3/4 = 1 (x-y/2-1/2)^2 + (3/4)(y^2-2y+1) = 1 (2x-y-1)^2 + 3(y-1)^2 = 4 只有二解 (2x-y-1)^2=4, (y-1)^2=0 或 (2x-y-1)^2 = 1, (y-1)^2 = 1 解聯立： 2x-y-1 = 2 1. { y-1 = 0 x=2, y=1 2x-y-1 =-2 2. { y-1 = 0 x=0, y=1 2x-y-1 = 1 3. { y-1 = 1 x=2, y=2 2x-y-1 = 1 4. { y-1 =-1 x=1, y=0 2x-y-1 =-1 5. { y-1 = 1 x=1, y=2 2x-y-1 =-1 6. { y-1 =-1 x=0, y=0 第二題令 x = u+2 (u+2)^3-6(u+2)^2+13(u+2)-2006 = 0 u^3+u-1996 = 0 ------(1) 令y = v-1 (v-1)^3+3(v-1)^2+4(v-1)+1998 = 0 v^3+v+1996 = 0 ------(2) (1)+(2)--->u^3+v^3+u+v = 0 (u+v)(u^2-uv+v^2) +(u+v) = 0 (u+v)(u^2-uv+v^2+1) = 0 u^2-uv+v^2+1 = (u-v/2)^2+(3/4)*v^2+1 > 0 故u+v= 0 x+y = u+2+v-1 = u+v+1 = 0+1 = 1 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 115.82.223.157

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haze

07/19 07:26, , 1^F

07/19 07:26, 1^F

※ 編輯: wettland5566 來自: 115.82.41.131 (07/19 08:16)

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wettland5566

07/19 08:17, , 2^F

07/19 08:17, 2^F

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iamduncan

07/19 16:46, , 3^F