初一 数学 因式分解 请详细解答,谢谢! (26 20:49:19)

来源:百度知道 编辑:UC知道 时间:2024/07/07 15:13:58
1.   计算[(3x-7)2-(x+5)2]/(4x-24)
2.  如果2x-y=10,求代数式[(x2+y2)-(x-y)2+2y(x-y)/4y的值
3.   证明两个连续奇数的平方差能被8整除
4.   计算(a2-b2)/(a-b)+(a2-2ab+b2)/(a-b)

1、[(3x-7)2-(x+5)2]/(4x-24)
=(3x-7+x+5)*(3x-7-x-5)/(4x-24)
=(4x-2)(2x-12)/(4x-24)
=2x-1

2、[(x2+y2)-(x-y)2+2y(x-y)/4y
=(x^2+y^2-x^2-y^2+2xy+2xy-2y^2)/4y
=(4xy-2y^2)/4y
=(2x-y)/2
=5

3、n>=1,(2n+1)^2-(2n-1)^2=(2n+1+2n-1)*(2n+1-2n+1)
=4n*2
=8n
所以能被8整除

4、(a2-b2)/(a-b)+(a2-2ab+b2)/(a-b)
=(a+b)(a-b)/(a-b)+(a-b)^2/(a-b)
=(a+b)+(a-b)
=2a

1.
[(3x-7)^2-(x+5)^2]/(4x-24)
=[(3x-7)-(x+5)][(3x-7)+(x+5)]/(4x-24)
=[(3x-7-x-5)(3x-7+x+5)]/(4x-24)
=[(2x-12)(4x-2)]/(4x-24)
=4[(x-6)(2x-1)]/[4(x-6)]
=2x-1

2.
[(x^2+y^2)-(x-y)^2+2y(x-y)]/4y
=[(x^2+y^2)-(x^2-2xy+y^2)+2xy-2y^2)]/4y
=[(x^2+y^2-x^2+2xy-y^2+2xy-2y^2)]/4y
=(4xy-2y^2)/4y
=2y(2x-y)/4y
=(2x-y)/2
=10/2
=5

3.
根据推导法得:
3^2-1^2=9-1=8
5^2-3^2=25-9=16
7^2-5^2=49-25=24
9^2-7^2=8