急、几道数学题、thank.(需要过程、听的懂就行)

3个回答

  • ①(x+y)^2(x-y)^2-(x-y)(x+y)(x^2+y^2)

    =[(x+y)(x-y)]^2-(x^2-y^2)(x^2+y^2)

    =[(x^2-y^2)]^2-(x^2-y^2)(x^2+y^2)

    =(x^2-y^2)[x^2-y^2-(x^2+y^2)]

    =-2(xy)^2+2y^4

    ②已知x^2+y^2+2x+4y+5=0,求2008x-2009y的值.

    由x^2+y^2+2x+4y+5=0得:

    x^2+2x+1+y^2+4y+4=0

    即(x+1)^2+(y+2)^2=0

    解得:x=-1,y=-2

    代入2008x-2009y可求出答案.

    ③1^2-2^2+3^2-4^2+…-98^2+99^2-100^2+101^2

    =(1+2)(1-2)+(3+4)(3-4)+…+(98+99)(98-99)+(100+101)(100-101)

    =(1+2)(-1)+(3+4)(-1)+…+(98+99)(-1)+(100+101)(-1)

    =(1+2+3+4+…+98+99+100+101)(-1)

    =-5151

    ④(2+1)(2^2+1)(2^4+1)…(2^2n+1)

    =(2-1)(2+1)(2^2+1)(2^4+1)…(2^2n+1)

    =(2^2-1)(2^2+1)(2^4+1)…(2^2n+1)

    =(2^4-1)(2^4+1)…(2^2n+1)

    =2^4n-1

    ⑤(2x-y)^3-(2x-y)^2(x-3y)

    =(2x-y)^2[2x-y-(x-3y)]

    =(2x-y)^2(x+y)

    ⑥当N为自然数时,代数式(n^2-n+1)(n^2-n+3)+1是一个完全平方式,请简要说明理由.

    (n^2-n+1)(n^2-n+3)+1

    =(n^2-n+1)(n^2-n+1)+2(n^2-n+1)+1

    =(n^2-n+1)^2+2(n^2-n+1)+1

    =(n^2-n+1+1)^2

    =(n^2-n+2)^2是完全平方式.

    ⑦已知x^2+x+1=0求x^3+2x^2+2006的值.