高手赐教，数学题，马上要过程和答案

显示全部楼层 · 2009-2-10 15:54:13

[（a-2/a^2+2a)-(a-1/a^2+4a+4)]/(a-4/a+2)其中a满足；a^2+2a-1=0

千问 · 2009-2-10 15:54:13

a^2+2a-1=0所以a^2+2a=1原式=[(a-2)/a(a+2)-(a-1)/(a+2)^2]*[(a+2)/(a-4)]=[(a-2)/a(a+2)]*[(a+2)/(a-4)]-[(a-1)/(a+2)^2]*[(a+2)/(a-4)]=(a-2)/[a(a-4)]-(a-1)/[(a+2)(a-4)]=[(a-2)(a+2)-a(a-1)]/[a(a+2)(a-4)]=(a^2-4-a^2+a)/[a(a+2)(a-4)]=(a-4)/[(a^2+2a)(a-4)]=1/(a^2+2a)=1/1=1...

千问 · 2009-2-10 15:54:13

[（a-2/a^2+2a)-(a-1/a^2+4a+4)]/(a-4/a+2)=[(a-2)(a+2)-a(a-1)]*(a+2)/a(a+2)^2(a-4)=(a-4)(a+2)/a(a+2)^2(a-4)=(a-4)/a(a+2)(a-4)=1/a(a+2)=1/(a^2+2a)=1很高兴为你解决问题！...

千问 · 2009-2-10 15:54:13

[（a-2/a^2+2a)-(a-1/a^2+4a+4)]/(a-4/a+2)=(a-2/a^2+2a-a+1/a^2-4a-4)/(a-4/a+2)=(-2a-1/a^2-4)/(a-4/a+2)=-(2a^3+4a^2+1)/(a^2+2a-4)=-[2a(a^2+2a)+1]/(1-4)=(2a+1)/3...