Question

这道题。求解( ｰ̀дｰ́ )初一数学题

Answer 1

由题意得{a+2=0
b-1/4=0
解得a=-2, b=1/4
原式=5a^2b-[2a^2b-ab^2+2a^2b-4]-2ab^2
=5a^2b-(4a^2b-ab^2-4)-2ab^2
=5a^2b-4a^2b+ab^2+4-2ab^2
=a^2b+4-ab^2
=(-2)^2×1/4+4-（-2）×（1/4）^2
=1+4+1/8
=5又1/8

Answer 2

由题意可知（a+2)²+|b-1/4|=0
又（a+2)²>=0,|b-1/4|>=0
所以a+2=0,b-1/4=0
得a=-2,b=1/4

化简所求式子
5a²b-[2a²b-(ab²-2a²b)-4]-2ab²=5a²b-[2a²b-ab²+2a²b-4]-2ab²=5a²b-[4a²b-ab²-4]-2ab²=5a²b-4a²b+ab²+4-2ab²=a²b-ab²+4=(-2)²×1/4-(-2)×(1/4)²+4=1+1/8+4=5+1/8=41/8

Answer 3

Answer 4