已知a=根号5 -1,那么:a+1=根号5
所以:
2a³+7a²-2a-12
=2a³+2a²+5a²+5a-7a-7-5
=2a²(a+1)+5a(a+1)-7(a+1)-5
=(a+1)(2a²+5a-7)-5
=(a+1)(2a+7)(a-1)-5
=(根号5)*(2根号5 - 2+7)*(根号5 -2)-5
=(5- 2根号5)(2根号5+5)-5
=5²-(2根号5)²-5
=25-20-5
=0
已知a=根号5 -1,那么:a+1=根号5
所以:
2a³+7a²-2a-12
=2a³+2a²+5a²+5a-7a-7-5
=2a²(a+1)+5a(a+1)-7(a+1)-5
=(a+1)(2a²+5a-7)-5
=(a+1)(2a+7)(a-1)-5
=(根号5)*(2根号5 - 2+7)*(根号5 -2)-5
=(5- 2根号5)(2根号5+5)-5
=5²-(2根号5)²-5
=25-20-5
=0