Question

一道数学题!急用!

一道数学题:
已知a^2(^=平方)-a-4=0,求4a^2-2(a^2-a+3)-0.5(a^2-a-4)-4a的值.
另外追加个解这道题的方法!

Answer 1

先化简
4a^2-2(a^2-a+3)-0.5(a^2-a-4)-4a
=1.5a^2-2.5a-4
a^2-a-4=0
a^2=a+4
=1.5(a+4)-2.5a-4
=-a+2
a^2-a-4=0,解出a,代入
好了

Answer 2

a^2-a-4=0
a^2=a+4
将a^2=a+4带入4a^2-2(a^2-a+3)-0.5(a^2-a-4)-4a就可以了

Answer 3

已知a^2-a-4=0

4a^2-2(a^2-a+3)-0.5(a^2-a-4)-4a
=4a^2-4a-2(a^2-a+3)-0.5(0)
=4(a^2-a)-2(a^2-a-4+3+4)
=4(a^2-a-4+4)-2(a^2-a-4+7)
=4(0+4)-2(0+7)
=16-14
=2

Answer 4

a^2-a-4=0
则a^2-a-4+7=7
即a^2-a+3=7
同时4*（a^2-a-4）=0
则4a^2-4a=16
代入
4a^2-2(a^2-a+3)-0.5(a^2-a-4)-4a
=4a^2-2*7-0.5*0-4a
=4a^2-4a-14
=16-14
=2