以知a-1/a=3,求a^4-1/a^4 的值
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解:由a-1/a=3两边平方得, a^2-2+1/a^2=9 ∴a^2+1/a^2=11 有a^2+2+1/a^2=13 ∴(a+1/a)^2=13 a+1/a=±√13 a^4-1/a^4 =(a^2+1/a^2)(a+1/a)(a-1/a) =11*(±√13)*3 =±33√13
题目不清楚: 若是(a-1)/a=3,求(a^4-1)/a^4 的值 则(a-1)/a=3,得1-(1/a)=3 所以,1/a=-2 (a^4-1)/a^4=1-(1/a^4)=1-(1/a)^4=1-(-2)^4=-15 若是a-(1/a)=3,求a^4-(1/a^4) 的值 则由a-(1/a)=3 a-(1/a)=3 a^2-1=3a [(a-(1/a)]^2=9 a^2-3a-1=0 a^2+(1/a)^2=11 得a=(3±√13)/2 得 (1/a)^2=11-a^2 由a^2-3a-1=0知a^2=3a+1=3(3±√13)/2 +1 (1/a)^2=11-[3(3±√13)/2 +1]=10-3(3±√13)/2 a^4-(1/a^4)=[a^2+(1/a)^2][a^2-(1/a)^2] =11*(±3√13) =±33√13。
a-1/a=3,也就是1-(1/a)=3 所以,1/a=-2 a^4-1/a^4=1-(1/a^4)=1-(1/a)^4=1-(-2)^4=-15
[a+(1/a)]^2 =[a-(1/a)]^2+4 =9+4 =13 所以 a+(1/a)=±√13 则 a^4-(1/a^4) =[a^2+(1/a^2)][a^2-(1/a^2)] ={[a-(1/a)]^2 +2}[a-(1/a)][a+(1/a)] =(9+2)×3×(±√13) =±33√13
答:a-1/a=3,9=(a-1/a)^2=a^2-2+1/a^2, a^2+1/a^2=11, a^2+2+1/a^2=13=(a+1/a)^2 a^4-1/a^...详情>>