已知实数a满足_已知实数a满足a^2

精选知识

题目是a^8+7a^-4

由a^2-a-1=0得,a^2=a+1,

于是a^4=(a+1)^2=a^2+2a+1=3a+2

a^8=(3a+2)^2=9a^2+12a+4=21a+13

a^8+7a^-4=21a+13+7/(3a+2)

=(63a^2+81a+26+7)/(3a+2)

=(63a+63+81a+33)/(3a+2)

=(144a+96)/(3a+2)=48

其他类似问题

问题1：已知实数a满足a^2-a-1=0则a^8+7a^-4的值为多少 求的值[数学科目]

a^8=[a^2]^4=(a+1)^4=(a^2+2a+1)^2=(a+1+2a+1)^2=(3a+2)^2=9a^2+12a+4=21a+13 a^4=[a^2]^2=(a+1)^2=3a+2 7a^-4=7/(3a+2) a^8+7a^-4=21a+13+7/(3a+2)

问题2：已知实数a满足a2-a-1=0．则a8+7a-4的值为______．[数学科目]

∵a²-a-1=0，
∴两边都除以a得，a-a^-1=1，
∴a²+a^-2=3，a⁴+a^-4=7，
∴a⁸+7a^-4，
=a⁴?a⁴+a⁴?a^-4-1+7a^-4，
=a⁴（a⁴+a^-4）+7a^-4-1，
=7a⁴+7a^-4-1，
=7×7-1，
=48．
故答案为：48．

问题3：已知实数a满足a²-a-1=0,则a八次方+7a负四次的值为?[数学科目]

a^2-a-1=0

a^2=a+1

a^4=(a+1)^2=a^+2a+1=3a+2

a^2-a=1

a-1=1/a

所以,原式=(3a+2)^2+7(a-1)^4

=9a^2+12a+4+7(a^2-2a+1)^2

=9a^2+12a+4+7(a+1-2a+1)^2

=9a^2+12a+4+7(a^2-4a+4)

=16(a^2-a+2)

=16(1+2)

=48

问题4：已知实数a,b满足a-7a+2=0,b-7b+2=0,求b/a+a/b的值如题[数学科目]

实数a、b满足条件a的平方—7a+2=0,b的平方—7b+2=0 说明a、b是方程x^2-7x+2=0的两根/得到 a+b=7,ab=2 b/a+a/b通分得到（a^2+b^2）/ab=((a+b)^2-2ab)/ab=45/2

问题5：已知实数a满足a的平方- a-1=0,求a的8次方+7a的负四次方[数学科目]

由a^2-a-1=0得,a^2=a+1,

于是a^4=(a+1)^2=a^2+2a+1=3a+2

a^8=(3a+2)^2=9a^2+12a+4=21a+13

a^8+7a^-4=21a+13+7/(3a+2)

=(63a^2+81a+26+7)/(3a+2)

=(63a+63+81a+33)/(3a+2)

=(144a+96)/(3a+2)=48