Question

已知(a-2)的平方+|b-1|=0,求代数式5ab-[2a^2b-(4ab^2-2a^2b)]

Answer 1

(a-2)的平方+|b-1|=0,
所以a-2=0;b-1=0;
a=2;b=1;
5ab-[2a^2b-(4ab^2-2a^2b)]
=10-[4-(8-8)]
=10-8
=2;