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=0④(2X+3)²=(X-1)² 1个回答 分类:数学 2014-10-29 问题解答: 我来补答 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=(X-1)²(2x+3)²-(x-1)²=0(2x+3+x-1)(2x+3-x+1)=0(3x+2)(x+4)=03x+2=0x=-2/3x+4=0x=-4 展开全文阅读