Factorization(4)

2008-09-30 4:22 am
1.(3x-2)^3-[2(2(x+1)]^3

2.a) (x+1)^2-(x-1)^2

2b) Hence by using (a) without using a calculator, factorize and evalute 2001^2-1999^2

3. 4x+5-12x^2

4.15x^2+22xy-56y^2

5. 250x^3+16y^3

6. x^2-x-2

回答 (2)

2008-09-30 7:04 am
✔ 最佳答案
sorry that I can't answer you the Q.5
1.(3x-2)^3-[2(2(x+1)]^3
=3^3x^3-2^3-[292^3x^3+1]
=3^3x^3-2^3-[16+2x^3+2]
=27x^3-8-16-2x^3-2
=25x^3-26

2.a) (x+1)^2-(x-1)^2 *.for this Q. , do you know a^2-b^2=(a+b)(a-b)?
=[(x+1)+(x-1)][(x+1)-(x-1)]
=x+1+x-1+x+1-x-1
=2x

2b) Hence by using (a) without using a calculator, factorize and evalute 2001^2-1999^2

2001^2-1999^2
=(2001+1999)(2001-1999)
=4000*2
=8000

3. 4x+5-12x^2
=4x-12x^2-15
=4x(1-3x)-15

4.15x^2+22xy-56y^2
=x(15x+22y)-56y^2

6. x^2-x-2
=x(x-1)-2
參考: ME
2008-09-30 6:01 am
1.
(3x-2)^3-[2(2(x+1)]^3
(27x^3-54x^2+36x-8)-(4x+4)^3
(27x^3-54x^2+36x-8)-(64x^3+192x^2+192x+64)
27x^3-54x^2+36x-8-64x^3-192x^2-192x-64
-37x^3-246x^2-156x-72

2a.
(x+1)^2-(x-1)^2
[(x+1)-(x-1)][(x+1)+(x-1)]
(2)(2x)
4x

2b.
2001^2-1999^2
(2000+1)^2-(2000-1)^2
4(2000)
8000

3.
4x+5-12x^2
-(12x^2-4x-5)
-(2x+1)(6x-5)

4.
15x^2+22xy-56y^2
(3x-4)(5x+14)

5.
250x^3+16y^3
2(125x^3+8y^3)
2(5x+2y)(25x^2-10xy+4y^2)

6.
x^2-x-2
(x-2)(x+1)


收錄日期: 2021-04-15 18:18:26
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