Question

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

Answer 1

①3X²+2X=0
x(3x+2)=0
x=0
或者3x+2=0
x=-2/3
②X(3X+2)-6(3X+2)=0
(3x+2)(x-6)=0
3x+2=0
x=-2/3
或者x-6=0
x=6
③9t²-(t-1)²=0
(3t+t-1)(3t-t+1)=0
(4t-1)(2t+1)=0
4t-1=0
t=1/4
或者2t+1=0
t=-1/2
④（2X+3）²=（X-1）²

(2x+3)²-(x-1)²=0
(2x+3+x-1)(2x+3-x+1)=0
(3x+2)(x+4)=0
3x+2=0
x=-2/3
x+4=0
x=-4

Answer 2

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