Question

若a-b=4 ab=1 求（-2ab+2a+3b)-(3ab+2b-2a)-(a+4b+ab)的值 帮忙解解好吗？

急~~~~~~~~~~

Answer 1

（－2ab+2a+3b）－（3ab+2b-2a）－（a+4b+ab）
=-2ab+2a+3b-3ab-2b+2a-a-4b-ab
=-6ab+3a-3b
=-6ab+3(a-b)
将a-b＝4，ab＝1代入上式
原式=-6*(1)+3*4=-6+12=6

所以（－2ab+2a+3b）－（3ab+2b-2a）－（a+4b+ab）＝6

Answer 2

6（-2ab+2a+3b)-(3ab+2b-2a)-(a+4b+ab)
=-2ab+2a+3b-3ab-2b+2a-a-4b-ab
=-6ab+3a-3b
=-6ab+3(a-b)
=-6*1+3*4
=-6+12
=6

Answer 3

6
要过程么~

Answer 4

（-2ab+2a+3b)-(3ab+2b-2a)-(a+4b+ab)
=-2ab+2a+3b-3ab-2b+2a-a-4b-ab
=-6ab+3a-3b
=-6ab+3(a-b)
=-6*1+3*4
=-6+12
=6