分解因式:(1).(x^2+2x+3)(x^2+2x+2)-30 (2).x^2+2xy+y^2-4x-7y-12 (3

3个回答

  • 1、原式=[(x²+2x)+3]×[(x²+2x)+2]-30

    =(x²+2x)²+5(x²+2x)+6-30

    =(x²+2x)²+5(x²+2x)-24

    =[(x²+2x)+8]×[(x²+2x)-3]

    =(x+3)(x-1)(x²+2x+8)

    2、原式=(x+y)²-4(x+y)-12=[(x+y)+2]×[(x+y)-6]=(x+y+2)(x+y-6)

    3、原式=-[(x-y)²-(x-y)-2]=-(x-y+1)(x-y-2)

    4、3x²-11x+10=0

    (3x-5)(x-2)=0

    x=5/3或x=2