已知a^2-3a+1=0求a^3/（a^6+1）的值.
∵a^2-3a+1=0
∴a^2+1=3a ……①
a^4=(3a-1)^2=9a^2-6a+1……②
3a=a^2+1……③
a^3/（a^6+1）
=a^3/[(a^2+1)(a^4-a^2+1)] 以①代入
=a^3/[3a(a^4-a^2+1)]
=a^2/[3(a^4-a^2+1)] 以②代入
=a^2/[3(8a^2-6a+2)] 以③代入
=a^2/{3[8a^2-2(a^2+1)+2]}
=a^2/[3(8a^2-2a^2-2+2)]
=1/24
∵A^2－3A＋1=0 ∴A＋1/A=3 ∴A^2＋2＋(1/A)^2=9 ∴A^2＋1/A^2=7 ∴A^3＋1/A^3 =(A＋1/A)(A^2－1＋1/A^2) =3×(7－1)=18 ∴A^3/(A^6＋1) =1/(A^3＋1/A^3) =1/18
已知a2-3a+1=0,显然a≠0,所以a+1/a=3,a^2+1/a^2=7 求a3/（a6+1)=1/(a^3+1/a^3)=1/( a+1/a)(a^2-1+1/a^2)=1/(3*6)=1/18
已知a2-3a+1=0求a3/（a6+1）的值. 解；由a^2-3a+1=0得a^2+1=3a a^3/(a^6+1)=a^3/(a^2+1)(a^4-a^2+1) =a^3/3a[(a^2+1)^2-3a^2] =a^2/3*[(3a)^2-3a^2] =a^2/3*6a^2 =1/18
答：a^2*b-2,b=2--->a^2=1 (-1/2)ab(a+a^3*b-a^5*b^2)? ? ? =(-1/2)(a^2*b+a^4*b^2-a^6*b^...详情>>
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