factorization的問題?

2007-09-17 12:30 am

請列識:
4)99p^2-(81q^2-p^2)
7)1+4u(1+u)
8)z^2+2-(2z+1)
20)16(p+q)^2-25q^2
22)75(c-3d)^2-27(2c-5d)^2
29)18t^2-57t-108
31)x^4-10x^2+9
34)(a+b)^3+(a-b)^3
32a)factorize 4a^2-4ab+b^2
b)use the result in (a) to factorize 4a^2-4ab+b^2-c^2
37)write down four possible values for c if x^2+cx+16
1a)(p^2+q^2)^2-4p^2q^2
b)36x^2-12x-169y^2+26y
2)it is given that a^2+b^2=39and ab=18.
hence find the value of a^4-2a^2b^2+b^4

回答 (1)

ok pan

2007-09-17 2:39 am

✔ 最佳答案

4.99p^2-(81q^2-p^2)
=99p^2-81q^2+p^2
=100p^2-81q^2
=(10p+9q)(10p-9q)
7.1+4u(1+u)
=1+4u+4u^2
=(1+2u)
8.z^2+2-(2z+1)
=z^2+2-2z-1
=z^2-2z+1
=(z-2)^2
20.16(p+q)^2-25q^2
=16p^2+32pq+16q^2-25q^2
=16p^2+32pq-9q^2
=(4p-q)(4p+9q)
22.75(c-3d)^2-27(2c-5d)^2
=75c^2-450cd+675d^2-108c^2+540cd-675d^2
=-33c^2+90cd
=-3c(11c-30d)
29.18t^2-57t-108
=3(2t-9)(3t+4)
31.x^4-10x^2+9
=(x^2-1)(x^2-9)
=(x+1)(x-1)(x+3)(x-3)
34.(a+b)^3+(a-b)^3
=a^3+3a^2*b+3ab^2+b^3+a^3-3a^2*b+3ab^2-b^3
=2a^3+6ab^2
=2a(a^2+3b^2)
32a.4a^2-4ab+b^2
=(2a-b)^2
b.4a^2-4ab+b^2-c^2
=(2a-b)^2-c^2
=(2a-b+c)(2a-b-c)
37. c=8,-8,10,-10,17,-17
1a.(p^2+q^2)^2-4p^2q^2
=p^4+2p^2q^2+q^4-4p^2q^2
=p^4-2p^2q^2+q^4
=(p^2-q^2)^2
b.36x^2-12x-169y^2+26y
=36x^2-169y^2-12x+26y
=(6x+13y)(6x-13y)-2(6x-13y)
=(6x-13y)(6x+13y-2)
2.a^4-2a^2b^2+b^4
=a^4+b^4-2a^2b^2
=(a^2+b^2)^2-2a^2b^2-2a^2b^2
=(a^2+b^2)^2-4a^2b^2
=(a^2+b^2)^2-4(ab)^2
=39^2-4(18)^2
=225

參考: 自己做

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