Question

提取公因式法(计算题)

1、2（x+y）^2-4(x+y)
2、1/2(x-2y)-(2y-x)^2
3、(2a+b)(a-2b)-(2b-a)(2a+3b)
4、-3a^2(x-2y)^3+12a(2y-x)^4
5、24a^2b^3c(x+y-z)-18a^3bc^2(z-y-x)-36abc(x-z+y)
6、3a^n(1-a)-2(a^n-a^n+1)(n是整数)
7、(a-b)^m-(b-a)^m+1(m是整数)
8、（x-y）(2p-q)-2(y-x)(2q-p)
要步骤

Answer 1

1、2（x+y）^2-4(x+y)
=2(x+y)²-2(x+y)*2
=2(x+y)*(x+y-2)

2、1/2(x-2y)-(2y-x)^2
=1/2(x-2y)-(x-2y)²
=1/2(x-2y)[1-2(x-2y)]
=1/2(x-2y)(-2x+4y+1)
=-1/2(x-2y)(2x-4y-1)

3、(2a+b)(a-2b)-(2b-a)(2a+3b)
=(2a+b)(a-2b)+(a-2b)(2a+3b)
=(a-2b)(2a+b+2a+3b)
=(a-2b)(4a+4b)
=4(a-2b)(a+b)

4、-3a^2(x-2y)^3+12a(2y-x)^4
=-3a²(x-2y)³+12a(x-2y)^4
=-3a(x-2y)³[a-4(x-2y)]
=-3a(x-2y)³(a-4x+8y)

5、24a^2b^3c(x+y-z)-18a^3bc^2(z-y-x)-36abc(x-z+y)
=24a²b³c(x+y-z)+18a³bc²(x+y-z)-36abc(x+y-x)
=6abc(x+y-z)(4ab²+3a²c-6)

6、3a^n(1-a)-2(a^n-a^n+1)(n是整数)
=3a^n*(1-a)-2a^n+2a^(n+1)
=a^n*[3(1-a)-2+2a]
=a^n*(3-3a-2+2a)
=a^n*(-a+1)
=-a^n*(a-1)

7、(a-b)^m-(b-a)^m+1(m是整数)
①m为偶数,则m+1为奇数
原式=(a-b)^m+(a-b)^(m+1)
=(a-b)^m*(1+a-b)
②m为奇数,则m+1为偶数
原式=(a-b)^m-(a-b)^(m+1)
=(a-b)^m*(1-a+b)

8、（x-y）(2p-q)-2(y-x)(2q-p)
=(x-y)(2p-q)+2(x-y)(2q-p)
=(x-y)*[(2p-q)+2(2q-p)]
=(x-y)(2p-q+4q-2p)
=(x-y)(3q)
=3q(x-y)