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

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  • 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)