唔係好難既數學題

設四個未知數為A 、B、C、D，
因為7.11=0.79×3×3
所以可以分以下幾種情況
1．其中一個數是0.79 則A+B+C=6.32
A×B×C=9
2．其中一個數是0.79×2 則 A+B+C=5.53
A×B×C=4.5
3．其中一個數是0.79×3 則 A+B+C=4.74
A×B×C=3
4．其中一個數是0.79×4 則A+B+C=3.95
A×B×C=2.25
5．其中一個數是0.79×5 則A+B+C=3.16
A×B×C=1.8
6．其中一個數是0.79×6 則A+B+C=2.37
A×B×C=1.5
……………………
其中第四中情況可能性最大，因為乘積的末位是5，所以和中必有5。
嘗試各種可能。
ex:2.25=0.25×9=0.75×3=0.15×15=……=1.25×1.8
而1.8必須滿足A×B=1.8
A+B=2.7 （3.95-1.25=2.7）
解得A=1.5 B=1.2
綜合以上解答 A 、B、C、D分別為3.16、1.25、1.5、1.2。4個未知數為3.16、1.25、1.5、1.2。

英王的教室（讓我教你們炮製一條好知識），http://hk.knowledge.yahoo.com/question/?qid=7006050200471的回答並不適用於這條數學題，就算如何修改都不適用。http://hk.knowledge.yahoo.com/question/?qid=7006050200471之所以要咁做，就是因為A, B, C, D是該四件物件的價錢，而物件的價錢只能是正數，頂多是兩位小數，佢之所以要咁做是可以理解的。

但係睇番呢題，這四個數的角色只是未知數，並不是該四件物件的價錢，所以並不存在正數、兩位小數等限制。這題只是一條聯立方程，不是應用題，故無須兼顧實用性。好似你的做法，就算數值上做得有幾啱都好，但是概念上你已經完全錯哂！！！

其實這題要咁做先至啱：

Let the four unknowns be a, b, c and d,
Then
　╭
＿│a + b + c + d = 7.11
　│abcd = 7.11
　╰
Let a = p , b = q , c = r ,for all p, q, r in C\{0}
Then d = 7.11 – p – q – r
∴The solution set of a + b + c + d = 7.11 is {(p , q , r , 7.11 – p – q – r) , p , q , r belongs to real number}
Put the solution set into abcd = 7.11 :
pqr(7.11 – p – q – r) = 7.11
7.11pqr – p²qr – pq²r – pqr² = 7.11
pqr² + p²qr + pq²r – 7.11pqr + 7.11 = 0
pqr² + (p²q + pq² – 7.11pq)r + 7.11 = 0
r = [– (p²q + pq² – 7.11pq) ± √[(– (p²q + pq² – 7.11pq))² – 4pq × 7.11]]/(2pq)
= [7.11pq – p²q – pq² ± √[(p²q + pq² – 7.11pq)² – 28.44pq]]/(2pq)
∴d = 7.11 – p – q – [7.11pq – p²q – pq² ± √[(p²q + pq² – 7.11pq)² – 28.44pq]]/(2pq)
= [14.11pq – 2p²q – 2pq² – 7.11pq + p²q + pq² (– or +) √[(p²q + pq² – 7.11pq)² – 28.44pq]]/(2pq)
= [7.11pq – p²q – pq² (– or +) √[(p²q + pq² – 7.11pq)² – 28.44pq]]/(2pq)

Note: p and q are independent of each other, i.e. they may be equal or unequal

Therefore, there are infinitely many solution of a, b, c and d.
Taking p = 1 and q = 2 as a case, we have
a = 1
b = 2
c = [7.11 × 1 × 2 – 1² × 2 – 1 × 2² ± √[(1² × 2 + 1 × 2² – 7.11 × 1 × 2)² – 28.44 × 1 × 2]]/(2 × 1 × 2)
= [8.22 ± √[(– 8.22)² – 56.88]]/4
= (8.22 ± √10.6844)/4
= (411/50 ± √(26711/2500))/4
= (411 ± √26711)/200
d = (411 (– or +) √26711)/200
　　╭
　　│a = 1
∴＿│b = 2
　　│c = (411 ± √26711)/200
　　│d = (411 (– or +) √26711)/200
　　╰
In fact the general expression is
　╭
　│a = p
＿│b = q
　│c = [7.11pq – p²q – pq² ± √[(p²q + pq² – 7.11pq)² – 28.44pq]]/(2pq)
　│d = [7.11pq – p²q – pq² (– or +) √[(p²q + pq² – 7.11pq)² – 28.44pq]]/(2pq)
　╰
for all p, q in C\{0}

Hence the four unknowns are p, q, [7.11pq – p²q – pq² + √[(p²q + pq² – 7.11pq)² – 28.44pq]]/(2pq) and [7.11pq – p²q – pq² – √[(p²q + pq² – 7.11pq)² – 28.44pq]]/(2pq) respectively, for all p, q in C\{0}.