6道初三用因式分解法解二元一次方程题 老大们帮帮忙

1个回答

  • ①169(x+3)²-196(x-2)²=0

    [13(x+3)]²-[14(x-2)]²=0

    [(13x+39)+(14x-28)][(13x+39)-(14x-28)]=0

    (27x+11)(-x+67)=0

    x1=-11/27

    x2=67

    ②4x²-3x-7=0

    (4x-7)(x+1)=0

    x1=7/4

    x2=-1

    ③4(x+3)²=9(2x-1)²

    [2(x+3)]²-[3(2x-1)]²=0

    [(2x+6)+(6x-3)][(2x+6)-(6x-3)]=0

    (8x+3)(-4x+9)=0

    x1=-3/8

    x2=9/4

    ④3(3-x)²+x(x-3)=0

    3(x-3)²+x(x-3)=0

    (x-3)[3(x-3)+x]=0

    (x-3)[3x-9+x]=0

    (x-3)[4x-9]=0

    x1=3

    x2=9/4

    ⑤(x+2)²+3(x+2)-4=0

    [(x+2)+4][(x+2)-1]=0

    (x+6)(x+1)=0

    x1=-6

    x2=-1

    ⑥(3x+2)(x-4)+11=0

    3x^2+2x-12x-8+11=0

    3x^2-10x+3=0

    (3x-1)(x-3)=0

    x1=1/3

    x2=3