Question

已知a+b=5,ab=3, 求2(a^2-b)-2(ab-b^2)-(a^2-3ab-7b)-2b^的值

Answer 1

2(a^2-b)-2(ab-b^2)-(a^2-3ab-7b)-2b^
展开得：
原式=a^2+ab+5b
=a（a+b）+5b
=5a+5b
=5(a+b)
=25

Answer 2

2a^2-2b-2ab+2b^2-a^2+3ab+7b+2b^2
=a^2+b^2+5b+2ab+3b^2-ab
=(a+b)^2-3+3b^2+5b
=25-3+3b^2+5b
(5-b)*b=3
b^2-5b+3=0
b=(10+-2√13)/4
5b=(50+-10√13)/4
b^2=(950+-250√13)/4
原式=22+[3(950+-250√13)+50+-10√13]/4
=22+-190√13+725
=747+-190√13

Answer 3

a+b=5
两边平方
a^2+2ab+b^2=25
a^2+b^2=25-2ab=19

a^3b-2a^2b^2+ab^3
=ab(a^2-2ab+b^2)
=3*(19-2*3)
=39