一道数学题,先化简再求值(a+b)(a-b)+(a+b)²-(6a²b-2ab²)+2b,其中a=-2,b=三分之一

Question

一道数学题,先化简再求值
(a+b)(a-b)+(a+b)²-(6a²b-2ab²)+2b,其中a=-2,b=三分之一

Answer 1

(a+b)(a-b)+(a+b)²-(6a²b-2ab²)+2b
=a²-b²+a²+2ab+b²-6a²b+2ab²+2b
=2a²+2ab-6a²b+2ab²+2b ,其中a=-2,b=三分之一
=2*(-2)²+2*(-2)*(1/3)-6*(-2)²*(1/3)+2*(-2)(1/3)²+2*1/3
=8-1/3-8-4/9+2/3
=1/3-4/9
=3/9-4/9
=-1/9