1. (x-3)[2次方]-4x[2次方]
2. a[2次方]-b[2次方]+2a[2次方]b-2ab[2次方]
3. x[4次方]-1
因式分解的數..
2006-11-19 7:38 pm
回答 (5)
2006-11-19 7:51 pm
✔ 最佳答案
以上題目都是考你運用:a(2次方)-b(2次方)=(a+b)(a-b)
1. (x-3)[2次方]-4x[2次方]
=(x-3+2x)(x-3-2x)
=(3x-3)(-x-3)
= -3(x-1)(x+3)
2.a[2次方]-b[2次方]+2a[2次方]b-2ab[2次方]
=(a+b)(a-b)+2ab(a-b)
=(a+b+2ab)(a-b)
3.x[4次方]-1
=(x[2次方])[2次方]-1
=(x[2次方]+1)(x[2次方]-1)
=(x[2次方]+1)(x+1)(x-1)
2006-11-19 8:05 pm
1. (x-3)^2-4x^2
= (x-3)(x-3)-4x^2
=x^2-6x+9-4x^2
= -3x^2 -6x+9
= (-3x-9)(x-1)
x=3or x=1
2. a^2-b^2+2a^2b-2ab^2
(not yet solve)
3. x^4-1
= (x^2-1)(x^2+1)
= (x-1)(x+1)(x^2+1)
x- 1 or -1
= (x-3)(x-3)-4x^2
=x^2-6x+9-4x^2
= -3x^2 -6x+9
= (-3x-9)(x-1)
x=3or x=1
2. a^2-b^2+2a^2b-2ab^2
(not yet solve)
3. x^4-1
= (x^2-1)(x^2+1)
= (x-1)(x+1)(x^2+1)
x- 1 or -1
2006-11-19 8:04 pm
上面第一位，你讀英文數，而唔知因式分解之意係咩，咁你咪答lo，同埋，我都係讀英文數，但我知因式分解英文係咩，係factorization，我知係我有去查果字，你冇，仲有係度講完一段野，最後咩都冇幫個問者。
1. (x-3)[2次方]-4x[2次方]
=(x-3)²- 4x²
=(x-3)²- (2x)²
=[(x-3)+(2x)][(x-3)-(2x)]
=(3x-3)(-x-3)
=3(x-3)(-x-3)
2. a[2次方]-b[2次方]+2a[2次方]b-2ab[2次方]
=a² - b² +2a²b - 2ab²
=(a+b)(a-b)+2ab(a-b)
=(a-b)(a+b+2ab)
3. x[4次方]-1
=(x²)²- (1²)²
=(x²)²- 1²
=(x²+1)(x²-1)
1. (x-3)[2次方]-4x[2次方]
=(x-3)²- 4x²
=(x-3)²- (2x)²
=[(x-3)+(2x)][(x-3)-(2x)]
=(3x-3)(-x-3)
=3(x-3)(-x-3)
2. a[2次方]-b[2次方]+2a[2次方]b-2ab[2次方]
=a² - b² +2a²b - 2ab²
=(a+b)(a-b)+2ab(a-b)
=(a-b)(a+b+2ab)
3. x[4次方]-1
=(x²)²- (1²)²
=(x²)²- 1²
=(x²+1)(x²-1)
2006-11-19 7:46 pm
1. 3(X+3)(1-X)
2. (a-b)(a+b+2ab)
3. (X[2次方]+1)(X+1)(X-1)
2. (a-b)(a+b+2ab)
3. (X[2次方]+1)(X+1)(X-1)
2006-11-19 7:42 pm
我有讀數....
但係.....英文....
唔知咩係因式分解....
好想幫你....但係無能為力...
你可以既話搵個英文出黎
我可以答你
但係.....英文....
唔知咩係因式分解....
好想幫你....但係無能為力...
你可以既話搵個英文出黎
我可以答你
收錄日期: 2021-04-22 00:52:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061119000051KK01523