adong1976
春芽
共回答了19个问题采纳率:89.5% 举报
原式=
[(1-1分之x+5)-(1-1分之x+4))+(1-1分之x+3)-(1-1分之x+2)]*[(x
+3)(x+5)](x^2+7x+13)
=[(1分之x+2)-(1分之x+3)+(1分之x+4)-(1分之x+5)]*)]*[(x
+3)(x+5)](x^2+7x+13)
=[(3分之(x+2)(x+5))-(1分之(x+3)(x+4))]*[(x
+3)(x+5)](x^2+7x+13)
=[3(x+3)](x+2)-(x+5x+4)](x^2+7x+13)
=[3(x+3)(x+4)-(x+2)(x+5)][(x+2)(x+4)](x^2+7x+13)
=[3(x^2+7x+12)-(x^2+7x+100][(x+2)(x+4)](x^2+7x+13)
=(2x^2+14x+26)[(x+2)(x+4)](x^2+7x+13)
=2[(x+2)(x+4)]
=(1分之x+2)-(1分之x+4)
请采纳我的答案,谢谢
1年前
4