Aa+Bb+Cab+D=0 and Ea+Fb+Gab+H=0－航空百大

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Aa+Bb+Cab+D=0 and Ea+Fb+Gab+H=0

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Aa+Bb+Cab+D=0 and Ea+Fb+Gab+H=0 where A,B,C,D,E,F,G,H are constant What is the value of a and b? 更新: 轉下問既方式: How to express a and b in terms of A,B,C,D,E,F,G,H? 更新 2: 做左個solution, 夠我自己用, 但以數學角度黎講呢個唔係完滿既答案. Aa+Bb+Cab+D=0------1 Ea+Fb+Gab+H=0------2 from 1, b=(-Aa-D)/(B+Ca)------3 sub. 3 into 2, Ea+F(-Aa-D)/(B+Ca)+Ga(-Aa-D)/(B+Ca)+D=0 更新 3: (BEa+CEa^2-AFa-DF-AGa^2-DGa+BD+CDa)/B+Ca=0 ((CE-AG)a^2+(BE-AF-DG+CD)a+(BD-DF))/B+Ca=0 Let X=CE-AG, Y=BE-AF-DG+CD, Z=BD-DF a=(-Y(+-)√Y^2-4XZ)/(2X) finally, sub. a into 3 to get b

最佳解答:

I have used your approach, and got about the same thing, except that I have the complete details for your reference. Since a has two solutions, so does b. So there are two sets of solutions, (a1,b1) and (a2,b2). Here's what I've got, and let me know if they agree with your solutions: a1=(-(B*(-sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F - 2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E)) / (2*B*G-2*C*F)-D)/((C*(-sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F-2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E))/(2*B*G-2*C*F) A) b1=(-sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F-2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E)/(2*B*G-2*C*F) a2=(-(B*(sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F-2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E)) / (2*B*G-2*C*F)-D)/((C*(sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F-2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E))/(2*B*G-2*C*F) A) b2=(sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F-2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E)/(2*B*G-2*C*F)

其他解答:

其實都差唔多, 雖然有少少錯. 用電腦計的答案係: a=(-Y(+-)√Y^2-4WZ)/(2X) X = C E - A G Y = B E - A F + D G - C H W = C F- B G Z = D E - A H 你自己對一對啦 2008-05-16 06:24:24 補充：係 a=(-Y(+-)√Y^2+4WZ)/(2X)|||||ABCDEFGH都係constant呀, 你所謂的unknown 只有a同b咋舉個例: Ax^2+Bx+C=0 x is: x=-(B(+-)√(B^2-4AC))/2A 我expect一個類似咁的答案... 我問果條問題Year 1時好似教過, 但我數學太差唔太記得點拆解|||||朋友你講緊笑啊?! 呢條都叫數?! 你唔好再多D unknown @@