1.factorize the followings
3(2x-3)²-7(2x-7)
2.a²-4ab+4b²-4
3.a²-4b²-a+2b
問3題數學題
2006-12-12 2:17 am
回答 (3)
2006-12-12 2:35 am
✔ 最佳答案
1) 3(2x-3)²-7(2x-7)=3(4x²-12x+9)-(14x-49)
=12x²-36x+27-14x+49
=12x²-50x+76
Therefore, it cannot be factorize.
2) a²-4ab+4b²-4
=a²-2(a)(2b)+(2b)²-4
=(a-2b)²-2²
=(a-2b+2)(a-2b-2)
3) a²-4b²-a+2b
=a²-(2b)²-(a-2b)
=(a+2b)(a-2b)-(a-2b)
=(a-2b)(a+2b-1)
參考: me
2006-12-12 2:35 am
2.a²-4ab+4b²-4
=(a-2b)²-4
=(a-2b-2)(a-2b+2)
3.a²-4b²-a+2b
=(a-2b)(a+2b)-(a-2b)
=(a-2b)(a+2b-1)
=(a-2b)²-4
=(a-2b-2)(a-2b+2)
3.a²-4b²-a+2b
=(a-2b)(a+2b)-(a-2b)
=(a-2b)(a+2b-1)
2006-12-12 2:24 am
第一題有冇寫錯題目，只factorize到一個factor 2
1. 3(2x-3)^2-7(2x-7)=3(4x^2-12x+9)-7(2x-7) =12x^2-36x+27-14x+49
=12x^2-50x+76 =2(6x^2-25x+38)
2. (a^2-4ab+4b^2)-4 =(a-2b)^2-(2)^2 = (a-2b+2)(a-2b-2)
3. a^2-(2b)^2 -(a-2b) =(a+2b)(a-2b) - (a-2b) = (a-2b)(a+2b-1)
(x+y)(x-y) 永遠相等 x^2-y^2
(x+y)^2 永遠相等 x^2+y^2+2xy
1. 3(2x-3)^2-7(2x-7)=3(4x^2-12x+9)-7(2x-7) =12x^2-36x+27-14x+49
=12x^2-50x+76 =2(6x^2-25x+38)
2. (a^2-4ab+4b^2)-4 =(a-2b)^2-(2)^2 = (a-2b+2)(a-2b-2)
3. a^2-(2b)^2 -(a-2b) =(a+2b)(a-2b) - (a-2b) = (a-2b)(a+2b-1)
(x+y)(x-y) 永遠相等 x^2-y^2
(x+y)^2 永遠相等 x^2+y^2+2xy
收錄日期: 2021-04-12 23:16:11
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