和 = 積 = 6.63

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.

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