分类：教育/科学-学习帮助
问题：两道数学计算题~~~初一滴~~很急啊~~~若a^2+a
若a^2+a-3=0,求a^4+2a^3-a-1的值已知a(a-1)-(a^2-b)=1，求式1/2(a^2+b^2)-ab的值
回答：若a^2+a-3=0,求a^4+2a^3-a-1的值 
a^2+a-3=0
a(a+1)=3
a^4+2a^3-a-1
=a^4+a^3+a^3-a-1
=a^3(a+1)+a(a^2-1)-1
=a^2*a(a+1)+a(a+1)(a-1)-1
=a^2*3+3*(a-1)-1
=3a^2+3a-3-1
=3a(a+1)-4
=3*3-4
=5

已知a(a-1)-(a^2-b)=1，求代数式1/2(a^2+b^2)-ab的值
a(a-1)-(a^2-b)=1
a^2-a-a^2+b=1
         b-a=1
           b=1+a
1/2(a^2+b^2)-ab
=1/2*(a^2+(1+a)^2)-a(1+a)
=1/2(a^2+1+2a+a^2)-a-a^2
=1/2(2a^2+2a+1)-a-a^2
=a^2+a+1/2-a-a^2
=1/2