遇到中二數學難題!

2008-06-17 1:47 am
把以下 轉成 ( 因式分解 )
1. a2 - 2ab + b2 - a +b -------------------------------- 注:a2 <--2次方
2. a (3b-c)+c-3b
3. x2 - 2x+1- 4y2 ------------------------------ 注:4y2 <--2為 2次方
4. (x2+4x+4)-(y-1)2 ------------------------- 注 : x2 / (y-1)2 2為 2次方
5. x2y + 2xy +y
6. x2y + 2xy +y - y3 ------------------- 5/6 注:x2y / y3 2次方/3次方
7. (a-b)2 - (a-b)2 ----------------------------------------注: 2----2次方

以上皆是問題
希請各位幫幫手 thx

回答 (2)

2008-06-17 2:18 am
✔ 最佳答案
其實我唔太明因式分解係將
x^2 - y^2 轉做 (x+y)(x-y)
定係
(x+y)(x-y)轉做x^2 - y^2
我見到你呢7題好似兩種形式都有...
1.
a^2 - 2ab + b^2 - a + b
= (a-b)^2 - (a-b)
= (a-b)(a-b) - (a-b)
= (a-b)(a-b-1)
註 a^2 姐係a既2次方 普通會考用計數機都會有呢個符號
以後係電腦入x既y次方可以咁表達

2.
a(3b-c)+c-3b
=3ab-ac+c-3b
=3ab-3b - ac+c
=3b(a-1) - c(a-1)------------------- 抽(a-1)呢個common factor出黎
=(3b-c)(a-1)

3.
x^2 - 2x+1- 4y^2
=(x-1)^2 - 4y^2
=(x-1)^2 - (2y)^2 --------- 常用分解用既公式x^2-y^2 = (x+y)(x-y)
=(x-1+2y)(x-1-2y) 你呢7題都用左好多咁既

4.
(x^2+4^x+4)-(y-1)^2------ 都係常用分解用公式(a+b)^2 =a^2+2ab-b^2
= (x+2)^2 - (y-1)^2 呢題只係調番轉咁解
= (x+2+y-1)[x+2-(y-1)]
= (x+y+1)(x-y+1)

5.
x^2y + 2xy +y
=y(x^2+2x+1)
=y(x+1)^2

6.
x^2y + 2xy +y - y^3
=y(x+1)^2-y^3 ---------呢題上半部同第5題一樣,so直接寫果part
=y[(x+1)^2-y^2] 既解法..略去左step
=y[(x+1+y)(x+1-y)]

7.(a-b)^2 - (a-b)^2
= 0 ---------兩個數一樣相減..唔駛計都知係0啦?定係打錯左?-.-

希望你明得哂呢7題既solution啦@@

2008-06-16 18:19:57 補充:
其實我唔太明因式分解係將
x^2 - y^2 轉做 (x+y)(x-y)
定係
(x+y)(x-y)轉做x^2 - y^2
我見到你呢7題好似兩種形式都有...

呢part唔駛理佢-.-做完我都明咩叫因式分解.事關我讀英中的..
參考: myself
2008-06-17 2:27 am
(1)

a^2 - 2ab + b^2 - a +b
=(a-b)^2-(a-b)
=(a-b)(a-b-1)

(2)

a (3b-c)+c-3b
=a (3b-c)-(3b-c)
=(3b-c)(a-1)

(3)

x^2-2x+1-4y^2
=(x-1)^2-4y^2
=(x-1)^2-(2y)^2
=(x-1+2y)(x-1-2y)【a^2-b^2=(a+b)(a-b)】

(4)

(x^2+4x+4)-(y-1)^2
=(x+2)^2-(y-1)^2
=(x+2+y-1)(x+2+y+1)【a^2-b^2=(a+b)(a-b)】

(5)

x^2y+2xy+y
=y(x^2+2x+1)
=y(x+1)^2

(6)

x^2y+2xy+y-y^3
=y(x+1)^2-y^3【From (5)】
=y〔(x+1)^2-y^2〕
=y(x+1+y)(x+1-y)【a^2-b^2=(a+b)(a-b)】

(7)

(a-b)^2-(a-b)^2
=0


For (7) if the first term is (a+b)^2, then

(a+b)^2-(a-b)^2
=(a+b+a-b)(a+b-a+b)【a^2-b^2=(a+b)(a-b)】
=(2a)(2b)
=4ab
參考: Me, hope it helps


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