由題意得：a^2-a-1=0
則a^2=a+1
由未達定理得：a+b=1
則原式=a^2*a+2b-5
=(a+1)*a+2b-5
=a^2+a+2b-5
=a+1+a+2b-5
=2a+2b-4
=2(a+b)-5
=2-5
=-3
在我回答的右上角點擊【采納答案】 那2a^2+5b^2的值為多少由題意得：a^2-a-1=0,b^2-b-1=0
則a^2=a+1,b^2=b+1
則2a^2+5b^2=2(a+1)+5(b+1)
=2a+1+5b+5
=2a+5b+6
=2(a+b)+3b+6
=2+3b+6
=8+3b