Question

a是方程x²+x-1/4=0的根,求a³-1/a的5次方+a的四次方－a³－a²

Answer 1

a是方程x2+x-1/4=0的根,则a^2+a-1/4=0 [^2指平方]a^2+a=1/4……(1)(a^3-1)/(a^5+a^4-a^3-a^2)=(a-1)(a^2+a+1)/(a^5-a^3+a^4-a^2)=(a-1)(a^2+a+1)/[a^3(a^2-1)+a^2(a^2-1)=(a-1)(a^2+a+1)/[(a^2-1)(a^3+a^2)]=(a-1)(a^2+a+1)/[(a+1)(a-1)*a^2*(a+1)]=(a^2+a+1)/[a(a+1)]^2 [由于a=1时方程并不成立,说明a≠1,从而a-1≠0,故分子分母同除以(a-1)]=(a^2+a+1)/(a^2+a)^2=(1/4+1)/(1/4)^2 [由上述(1)知,a^2+a=1/4,代入可得此结论]=(5/4)/(1/16)=(5/4)*16=20