數-代數 (15) - a(a-b)

2009-10-09 10:54 pm
a(a-b)=aa-ab
a(a-b)=b(b+a)
ab=b(b+b)
a=?
b=?
a(a-b)=b(b+a)
=>aa-ab=bb+ab
ab=b(b+b)
=>a=b+b
=>a=2b
aa-ab=bb+ab
4bb-2bb=bb+2bb
=>4-2=1+2
This is wrong , but can you to make that right(Hint : you can try the zero)
If no,say!
更新1:

a don't = b

更新2:

good

回答 (2)

2009-10-09 11:40 pm
✔ 最佳答案
a(a - b) = b(b + a) ... (1)
ab = b(b + b) ... (2)
ab = b(2b)
ab - 2bb = 0
b(a - 2b) = 0
a = 2b或b = 0
情況 (1) a = 2b
(1) => 2b(2b - b) = b(b + 2b)
4bb - 2bb = bb + 2bb
2bb = 3bb
0 = 3bb - 2bb
bb = 0
b = 0
a = 2b = 0
情況 (2) b = 0
(1) => a(a - 0) = 0(0 + a)
aa = 0
a = 0
所以兩種情況結論都是a = b = 0
題目中提到4bb - 2bb = bb + 2bb簡化為4 - 2 = 1 + 2是建基於b不等於0的假設,因此不成立。
若題目要求a不等於b, 則此題沒解。
2009-10-10 1:17 am
a(a - b) = b(b + a) ... (1)

ab = b(b + b) ... (2)

ab = b(2b)

ab - 2bb = 0

b(a - 2b) = 0

a = 2b或b = 0

情況 (1) a = 2b

(1) => 2b(2b - b) = b(b + 2b)

4bb - 2bb = bb + 2bb

2bb = 3bb

0 = 3bb - 2bb

bb = 0

b = 0

a = 2b = 0

情況 (2) b = 0

(1) => a(a - 0) = 0(0 + a)

aa = 0

a = 0

所以兩種情況結論都是a = b = 0

題目中提到4bb - 2bb = bb + 2bb簡化為4 - 2 = 1 + 2是建基於b不等於0的假設,因此不成立。

若題目要求a不等於b, 則此題沒解。


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