问题
填空题
已知a=
|
答案
由已知得（a+1）2=5，所以a2+2a=4
则原式=2a3+4a2+2a+3a2-4a-12
=2a（a2+2a+1）+3a2-4a-12
=2a（a+1）2+3a2-4a-12
=2a×5+3a2-4a-12
=3a2+6a-12
=3（a2+2a）-12
=3×4-12
=0
故答案0
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已知a=
|
由已知得（a+1）2=5，所以a2+2a=4
则原式=2a3+4a2+2a+3a2-4a-12
=2a（a2+2a+1）+3a2-4a-12
=2a（a+1）2+3a2-4a-12
=2a×5+3a2-4a-12
=3a2+6a-12
=3（a2+2a）-12
=3×4-12
=0
故答案0