提取公因式法(计算题)1、2(x+y)^2-4(x+y)2、1/2(x-2y)-(2y-x)^23、(2a+b)(a-2
1个回答

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)