用因式分解法解下列二元一次方程①3X²+2X=0②X(3X+2)-6(3X+2)=0③9t-(t-1)²=0④（2X+3）²=（X-1）²
问题描述：
用因式分解法解下列二元一次方程①3X²+2X=0②X(3X+2)-6(3X+2)=0③9t-(t-1)²=0
④（2X+3）²=（X-1）²
答
1、x(3x+2)=0,则x=0或x=-2/3
2、(x-6)(3x+2)=0,则x=6或x=-2/3
3、9t-(t^2-2t+1)=9t-t^2+2t-1=-t^2+11t-1=0
t^2-11t+1=0
用配方法：t^2-2*11/2t+(11/2)^2-(11/2)^2+1=0
(t-11/2)^2=117/4
t-11/2=正负（3根号13）/2
t=(3根号13+11）/2或（11-3根号13）/2
4、4x^2+12x+9=x^2-2x+1
3x^2+14x+8=0
用十字交叉法：（3x+2)(x+4)=0
x=-2/3或-4
答
①3X²+2X=0x(3x+2)=0x=0或者3x+2=0x=-2/3②X(3X+2)-6(3X+2)=0(3x+2)(x-6)=03x+2=0x=-2/3或者x-6=0x=6③9t²-(t-1)²=0(3t+t-1)(3t-t+1)=0(4t-1)(2t+1)=04t-1=0t=1/4或者2t+1=0t=-1/2④（2X+3）²=...