已知a-b=5,ab=-1,求(2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)的值

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已知a-b=5,ab=-1,求(2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)的值
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已知a-b=5,ab=-1,求(2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)的值
已知a-b=5,ab=-1,求(2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)的值

已知a-b=5,ab=-1,求(2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)的值
(2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)
=2a+3b-2ab-a-4b-ab-3ab-2b+2a
=3a-3b-5ab=3(a-b)-6ab=15-(-6)=21

(2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)
=2a+3b-2ab-a-4b-ab-3ab-2b+2a
=3a-3b-6ab
=3(a-b)-6ab
=3*5-6*(-1)
=15+6
=21

是21

(2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)=3a-3b-6ab=3(a-b)-6ab=3*5-6*(-1)=21

答案是21

化简不就行了??

由已知可得:
(2a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)
=2a+3b-2ab-a-4b-ab-3ab-2b+2a
=3a-3b-6ab
=3(a-b)-6ab
=3×5-6×(-1)
=21