已知(a+1)^2+|b-2|=0,
a+1=0;
a=-1;
b-2=0;
b=2;
求多項(xiàng)式a^2+3ab-7a^2b^2-2ab+1+5a^2b^2的值
=a²+ab-2a²b²+1
=(-1)²+(-1)×2-2×(-1)²×2²+1
=1-2-8+1
=-8;
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