计算：a^2-16/（a+4)(a-2)除以（a-4)a^2+4-4a/a-2;若x-1/x=3,求x^2+1/x^2=()x^4+1/x^4=()

如题所述

第1个回答 2012-03-20

a^2-16/（a+4)(a-2)除以（a-4)/a^2+4-4/a-2
=(a-4)(a+4)/(a+4)(a-2)÷[(a-4)/a²]-4/(a-2)
=a²/(a-2)-4/(a-2)
=(a²-4)/(a-2)
=(a+2)(a-2)/(a-2)
=a+2

若x-1/x=3,求x^2+1/x^2=() x^4+1/x^4=*()
x-1/x=3
(x-1/x)²=9
x²-2+1/x²=9
∴x²+1/x²=11

x²+1/x²=11
(x²+1/x²)=121
x^4+2+1/x^4=121
∴x^4+1/x^4=119

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计算：a^2-16/（a+4)(a-2)除以（a-4)*a^2+4-4a/a-2;若x-1/x=3,求x^2+1/x^2=()x^4+1/x^4=*()

计算：a^2-16/（a+4)(a-2)除以（a-4)a^2+4-4a/a-2;若x-1/x=3,求x^2+1/x^2=()x^4+1/x^4=()