因式分解...

Question

以下兩條因式分解點計? 請列式...

1) 12v - 12v^2 - 6u + 3u^2
2) 3/5(a+b)^2 - 5/3(a-b)^2

Answer 1

以下兩條因式分解點計? 請列式...
1) 12v - 12v2 - 6u + 3u2
12v - 12v2 - 6u + 3u2
= 3(u2 - 4v2) + 6(2v - u)
= 3[u2 - (2v)2] + 6(2v - u)
= 3(u - 2v)(u + 2v) - 6(u - 2v) 【應用恆等式 x2-y2=(x-y)(x+y)】
= 3(u - 2v)[(u + 2v) - 2] 【抽出 3(u-2v)】
= 3(u - 2v)(u + 2v - 2)
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2) 3/5(a+b)2 - 5/3(a-b)2
我在你的題目加上了括號令它更易被了解，希望我的理解沒錯。
(3/5)(a+b)2 - (5/3)(a-b)2
= 9(a+b)2/15 - 25(a-b)2/15 【通分母】
= (1/15)[9(a+b)2 - 25(a-b)2]
= (1/15)[(3(a+b))2 - (5(a-b))2]
= (1/15)[3(a+b) - 5(a-b)][3(a+b) + 5(a-b)] 【應用恆等式 x2-y2=(x-y)(x+y)】
= (1/15)(3a + 3b - 5a + 5b)(3a + 3b + 5a - 5b)
= (1/15)(-2a + 8b)(8a - 2b)
= (4/15)(4b-a)(4a-b)

= 4(4b-a)(4a-b)/15

2007-01-26 18:35:52 補充：
小小補充：factorization 常用的公式有：a²-b² = (a-b)(a+b)(a+b)² = a²+2ab+b²(a-b)² = a²-2ab+b²只是不用的項代入 a 和 b。

Answer 2

1) 12v - 12v^2 - 6u + 3u^2
=3(4v-4v^2-2u+u^2)
=3[(u+2v)(u-2v)-2(u-2v)]
=3(u+2v-2)(u-2v)


2) 3/5(a+b)^2 - 5/3(a-b)^2

=[9(a-b)^2-25(a+b)^2]/15(a+b)^2(a-b)^2--------通分~
=[3(a-b)+5(a+b)][3(a-b)-5(a+b)]/15(a+b)^2(a-b)^2
=(3a-3b+5a+5b)(3a-3b-5a-5b)/15(a+b)^2(a-b)^2
=(8a+2b)(-2a-8b)/15(a+b)^2(a-b)^2
=-4(4a+b)(a-4b)/15(a+b)^2(a-b)^2

Answer 3

2.
Not sure if you have your question right.
As you presented it:
3/5(a+b)^2-5/3(a-b)^2
=(sqrt(3/5)(a+b)+sqrt(5/3)(a-b)) ( sqrt(3/5)(a+b)-sqrt(5/3)(a-b))

If the question had been 3/5(a+b)^2-3/5(a-b)^2, then
3/5(a+b)^2-3/5(a-b)^2
=3/5((a+b)^2-(a-b)^2)
=3/5((a+b)+(a-b))((a+b)-(a-b))
=3/5(2a)(2b)
=12ab/5

Answer 4

1) 12v - 12v^2 - 6u + 3u^2
=12(1-v)-3u(2-u)