F.2數學,唔該幫幫手呀!急急急...(20點)

2010-10-02 12:59 am
1.Expand the following:

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

2.Factorize each of the following:

a) 4x^2-(y^2-2y+1)

b) 1-u^2+6uv-9v^2

c) a^2(a-1)^2+(a-1)^2+2a(1-a)^2

d) (x^2-x)^2+1/2(x^2-x)+1/16

唔該唔好用cross-method黎計factorization
淨係可以用take out common factor,
grouping terms同use identities

吾該,幫幫手呀:)
thxxxxxx;)

回答 (5)

2010-10-02 1:20 am
✔ 最佳答案
1. (a-5)^2-(b+3)^2
= (a-5-3-b)(a-5+b+3)
= (a-b-8)(a+b-2)
= a^2 +ab -2a -ab -b^2 + 2b -8a-8b+16
= a^2 - b^2 - 10 a - 6 b + 16

2a. 4x^2-(y^2-2y+1)
= 4x^2 - (y-1)^2
= (2x+y-1)(2x-y+1)

b. 1-u^2+6uv-9v^2
= 1-(u^2-6uv+9v^2)
= 1-(u-3v)^2
= (1-u+3v)(1+u-3v)

c. a^2(a-1)^2+(a-1)^2+2a(1-a)^2
= a^2(a-1)^2 +2a(a-1)^2 + (a-1)^2 .... because ( (a-1)^2 = (1-a)^2 )
= ( a(a-1) + (a-1) )^2
= (a^2 -a + a - 1)^2
= (a^2-1)^2
= (a+1)^2 (a-1)^2

d. (x^2-x)^2+1/2(x^2-x)+1/16
= (x^2-x + 1/4)^2
=( (x-1/2)^2 )^2
= (x - 1/2)^4
參考: gonkii.co.cc 網上問功課 (晚上11:00後關閉)
2010-10-02 1:24 am
我早早就把答案寄到你電郵信箱了~~希望幫到你!
2010-10-02 1:24 am
(a-5)^2-(b+3)^2
a^2-10a+25-b^2-6b-9
a^2-10a-6b+16
_________________
4x^2-(y^2-2y+1)
4x^2-(y-1)^2
(2x-y+1)(2x+y-1)
______________
1-u^2+6uv-9v^2
1-(u^2-6uv+9v^2)
1-(u-3v)^2
(1-u+3v)(1+u-3v)
_______________
a^2(a-1)^2+(a-1)^2+2a(1-a)^2
a^2(a-1)^2+(a-1)^2+2a(a-1)^2
(a-1)^2(a^2+1+2a)
(a-1)^2(a+1)^2
________________
(x^2-x)^2+1/2(x^2-x)+1/16
(1/16)(16(x^2-x)^2+8(x^2-x)+1))
(1/16)((4x^2-4x)^2+8(x^2-x)+1))
(1/16)(4x^2-4x-1)^2
2010-10-02 1:12 am
(a-5)^2-(b+3)^2
=a^2-5^2-b^2+3^2
=a^2-25-b^2+9
=a^2-16-b^2
2010-10-02 1:05 am
e d數唔係應該中四先交既咩?


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