標題:
和 = 積 = 6.63
發問:
最佳解答:
1. x=4.08 y=1.25 z=1.3 xyz=6.63 x+y+z=6.63 2. a=5 b=-0.85 c=-0.52 d=3 abcd=6.63 a+b+c+d=6.63 2007-03-21 06:44:45 補充： Overlooked following solution(不用負數), here it is:1.25 1.25 2.21 1.92=6.631.25*1.25*2.21*1.92=6.63 2007-03-21 06:52:55 補充： The problem can be converted to:A B C D=663A*B*C*D=663000000where A,B,C,D are whole numbers.The factors of 663000000 are 2^6,3,5^6,13,17So by testing numbers that are made up of these factors, there are not many to try. 2007-03-21 06:57:50 補充： It is important to limit the solution to integers as above, as an infinite set of real (decimal) solutions exist, but they violate the condition:不可4捨5入.
其他解答:
很好的方法。|||||1.) ╭ ＿│x + y + z = 6.63 │xyz = 6.63 ╰ Let x = s , y = t , where s , t are any real number Then z = 6.63 – s – t ∴The solution set of x + y + z = 6.63 is {(s , t , 6.63 – s – t) , s , t belongs to real number} Put the solution set into xyz = 6.63 : st(6.63 – s – t) = 6.63 6.63st – s2t – st2 = 6.63 st2 + (s2 – 6.63s)t + 6.63 = 0 t = [– (s2 – 6.63s) ± √((– (s2 – 6.63s))2 – 4s × 6.63))]/(2s) = [6.63s – s2 ± √(s^4 – 13.26s3 + 43.9569s2 – 26.52s)]/(2s) ╭ │x = s ∴─┤y = [6.63s – s2 (+ or –) √(s^4 – 13.26s3 + 43.9569s2 – 26.52s)]/(2s) │z = [6.63s – s2 (– or +) √(s^4 – 13.26s3 + 43.9569s2 – 26.52s)]/(2s) ╰ Where s^4 – 13.26s3 + 43.9569s2 – 26.52s ≧ 0 and s > 0 and s ≠ 1 and [6.63s – s2 ± √(s^4 – 13.26s3 + 43.9569s2 – 26.52s)]/(2s) > 0 and [6.63s – s2 ± √(s^4 – 13.26s3 + 43.9569s2 – 26.52s)]/(2s) ≠ 1 2.) ╭ ＿│a + b + c + d = 6.63 │abcd = 6.63 ╰ Let a = p , b = q , c = r , where p , q , r are any real number Then d = 6.63 – p – q – r ∴The solution set of a + b + c + d = 6.63 is {(p , q , r , 6.63 – p – q – r) , p , q , r belongs to real number} Put the solution set into abcd = 6.63 : pqr(6.63 – p – q – r) = 6.63 6.63pqr – p2qr – pq2r – pqr2 = 6.63 pqr2 + (p2q + pq2 – 6.63pq)r + 6.63 = 0 r = [– (p2q + pq2 – 6.63pq) ± √((– (p2q + pq2 – 6.63pq))2 – 4pq × 6.63))]/(2pq) = [6.63pq – p2q – pq2 ± √((p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq)]/(2pq) ∴a = p b = q c = [6.63pq – p2q – pq2 (+ or –) √((p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq)]/(2pq) d = [6.63pq – p2q – pq2 (– or +) √((p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq)]/(2pq) Where (p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq ≧ 0 and p > 0 and p ≠ 1 and q > 0 and q ≠ 1 and [6.63pq – p2q – pq2 ± √((p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq)]/(2pq) > 0 and [6.63pq – p2q – pq2 ± √((p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq)]/(2pq) ≠ 1|||||1. x=4.08 y=1.25 z=1.3 xyz=6.63 x+y+z=6.63 2. a=5 b=-0.85 c=-0.52 d=3 abcd=6.63 a+b+c+d=6.63|||||和 = 積 = 6.63 x + y + z = x * y * z = 6.63 a + b + c + d = a * b * c * d = 6.63 1.) 求 x,y,z 2.) 求 a,b,c,d 條件: 數字可以重覆 不可以是 0 和 1 不可4捨5入 按照你以上的條件，笞案如下： 1) x=4.08 y=1.25 z=1.3 xyz=6.63 x+y+z=6.63 2) a=5 b=-0.85 c=-0.52 d=3 abcd=6.63 a+b+c+d=6.63
和 = 積 = 6.63
發問:
此文章來自奇摩知識+如有不便請留言告知
x + y + z = x * y * z = 6.63 a + b + c + d = a * b * c * d = 6.63 1.) 求 x,y,z 2.) 求 a,b,c,d 條件: 數字可以重覆 不可以是 0 和 1 不可4捨5入 更新: 多謝兩位, 特別y6972593, 是答得相當不錯, 而且非常快。 其實問題是不可以用負數, 題目並沒有說清楚, 抱歉! 請再嘗試! 更新 2: paul2357, 一大堆equations, 即是計唔計倒呀! 如果可以用equations直接計得出 a, b, c,d, 就真係好勁! 我要嘅答案是 a=?; b=?; c=?; d=?最佳解答:
1. x=4.08 y=1.25 z=1.3 xyz=6.63 x+y+z=6.63 2. a=5 b=-0.85 c=-0.52 d=3 abcd=6.63 a+b+c+d=6.63 2007-03-21 06:44:45 補充： Overlooked following solution(不用負數), here it is:1.25 1.25 2.21 1.92=6.631.25*1.25*2.21*1.92=6.63 2007-03-21 06:52:55 補充： The problem can be converted to:A B C D=663A*B*C*D=663000000where A,B,C,D are whole numbers.The factors of 663000000 are 2^6,3,5^6,13,17So by testing numbers that are made up of these factors, there are not many to try. 2007-03-21 06:57:50 補充： It is important to limit the solution to integers as above, as an infinite set of real (decimal) solutions exist, but they violate the condition:不可4捨5入.
其他解答:
很好的方法。|||||1.) ╭ ＿│x + y + z = 6.63 │xyz = 6.63 ╰ Let x = s , y = t , where s , t are any real number Then z = 6.63 – s – t ∴The solution set of x + y + z = 6.63 is {(s , t , 6.63 – s – t) , s , t belongs to real number} Put the solution set into xyz = 6.63 : st(6.63 – s – t) = 6.63 6.63st – s2t – st2 = 6.63 st2 + (s2 – 6.63s)t + 6.63 = 0 t = [– (s2 – 6.63s) ± √((– (s2 – 6.63s))2 – 4s × 6.63))]/(2s) = [6.63s – s2 ± √(s^4 – 13.26s3 + 43.9569s2 – 26.52s)]/(2s) ╭ │x = s ∴─┤y = [6.63s – s2 (+ or –) √(s^4 – 13.26s3 + 43.9569s2 – 26.52s)]/(2s) │z = [6.63s – s2 (– or +) √(s^4 – 13.26s3 + 43.9569s2 – 26.52s)]/(2s) ╰ Where s^4 – 13.26s3 + 43.9569s2 – 26.52s ≧ 0 and s > 0 and s ≠ 1 and [6.63s – s2 ± √(s^4 – 13.26s3 + 43.9569s2 – 26.52s)]/(2s) > 0 and [6.63s – s2 ± √(s^4 – 13.26s3 + 43.9569s2 – 26.52s)]/(2s) ≠ 1 2.) ╭ ＿│a + b + c + d = 6.63 │abcd = 6.63 ╰ Let a = p , b = q , c = r , where p , q , r are any real number Then d = 6.63 – p – q – r ∴The solution set of a + b + c + d = 6.63 is {(p , q , r , 6.63 – p – q – r) , p , q , r belongs to real number} Put the solution set into abcd = 6.63 : pqr(6.63 – p – q – r) = 6.63 6.63pqr – p2qr – pq2r – pqr2 = 6.63 pqr2 + (p2q + pq2 – 6.63pq)r + 6.63 = 0 r = [– (p2q + pq2 – 6.63pq) ± √((– (p2q + pq2 – 6.63pq))2 – 4pq × 6.63))]/(2pq) = [6.63pq – p2q – pq2 ± √((p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq)]/(2pq) ∴a = p b = q c = [6.63pq – p2q – pq2 (+ or –) √((p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq)]/(2pq) d = [6.63pq – p2q – pq2 (– or +) √((p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq)]/(2pq) Where (p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq ≧ 0 and p > 0 and p ≠ 1 and q > 0 and q ≠ 1 and [6.63pq – p2q – pq2 ± √((p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq)]/(2pq) > 0 and [6.63pq – p2q – pq2 ± √((p^4)q2 + 2p3q3 + p2(q^4) – 13.26p3q2 – 13.26p2q3 + 43.9569p2q2 – 26.52pq)]/(2pq) ≠ 1|||||1. x=4.08 y=1.25 z=1.3 xyz=6.63 x+y+z=6.63 2. a=5 b=-0.85 c=-0.52 d=3 abcd=6.63 a+b+c+d=6.63|||||和 = 積 = 6.63 x + y + z = x * y * z = 6.63 a + b + c + d = a * b * c * d = 6.63 1.) 求 x,y,z 2.) 求 a,b,c,d 條件: 數字可以重覆 不可以是 0 和 1 不可4捨5入 按照你以上的條件，笞案如下： 1) x=4.08 y=1.25 z=1.3 xyz=6.63 x+y+z=6.63 2) a=5 b=-0.85 c=-0.52 d=3 abcd=6.63 a+b+c+d=6.63
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