谁能帮我解下列一元二次方程啊(4x+3)(5-X)=02X^2-4X-5=03x(x-1)=2-2xx(x+6)=7(x

2个回答

  • 1,(4x+3)(5-X)=0→→4x×5-4x×x+3×5-3×x=0,即20X-4X²+15-3X=0,→→

    4X²-17X-15=0→→﹙4X+3﹚﹙X-5﹚=0,所以4X+3=0或X-5=0

    即X=-3/4或X=5

    2,2X^2-4X-5=0→→

    3,3x(x-1)=2-2x→→3X²-3X+2X-2=3X²-X-2=0→→﹙3X+2﹚﹙X-1﹚=0,所以

    3X+2=0或X-1=0,即X=-2/3或X=1

    4,x(x+6)=7→→X²+6X-7=0→→﹙X+7﹚﹙X-1﹚=0,所以X+7=0或X-1=0,即

    X=-7或X=1

    5,(x-1)^2+2x(x-1)=0→→X²+1-2X+2X²-2X=3X²-4X+1=0→→

    ﹙3X-1﹚﹙X-1﹚=0,所以3X-1=0或X-1=0,即

    X=1/3或X=1

    6,-3x^2-4x+4=0→→3X²+4X-4=0→→﹙3X-2﹚﹙X+2﹚=0,所以

    3X-2=0或X+2=0,即X=2/3或X=-2

    7,(X-1)(X+2)=70→→X²+2X-X-2-70=X²+X-72=0→→﹙X+9﹚﹙X-8﹚=0,

    所以X+9=0或X-8=0,即X=-9或X=8

    8,(3-X)^2+X^2→→