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发表于 2007-8-9 18:10:42
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第(6)题解答
α(α+1)+β(β+1)=(α+1)(β+1)% c' U: ~; h, R1 t! q; o
α^2+α+β^2+β=αβ+α+β+1
* @, M- C T* N+ Eα^2+β^2+2αβ=3αβ+1
$ H2 t+ E5 M3 @# i B9 y0 {(α+β)^2=3αβ+1
1 y' F( D7 V% t {' w' b% f将α+β=-(A+B),αβ=4AB/3代入
& v/ @0 W9 D3 k7 i+ h(A+B)^2=4AB+1----------------------------------------------------------------(1)
; Q, R1 j- v3 M2 L+ T8 b/ Y8 O(A-B)^2=1,A>B,A-B=1--------------------------------------------------------(2)
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7 u7 W- V/ E4 J% s& E3 ], \" N方程有两实根,[3(A+B)]^2-4*3*4AB>=0+ C& ?# ~- n* N, z
9(A+B)^2-48AB>=0" z2 R1 ]; H- h' ?4 a l1 F* ^
代入(1)得 36AB+9-48AB>=0
7 {- N6 v3 |3 P AB<=3/4-------------------------------------(3)# L* ^# q! F5 X8 T
因为A,B均为整数,根据条件(2),(3)可穷举出满足此条件的只有A=1,B=0;A=0,B=-1.
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