6道初三用因式分解法解二元一次方程题 老大们帮帮忙

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