Question

一道数学题，急急急

已知a²-3a+1=0，求a³/（a的六次方+1）的值。【详细解释哦】

Answer 1

解:a²-3a+1=0
则 a²=3a-1 a很显然不等于0
所以 要想 求a³/a^6+1
我们不妨先求其倒数
即 (a^6+1)/a³
=a³+1/a³ (立方和公式)
=(a+1/a)(a²-1+1/a²) (通分)
=(a²+1)/a x (a²-1+1/a²) (将a²=3a-1 带入)
=3a/a x [(3a-1)-1+1/(3a-1)] (后面的通分)
=3 x [(3a-1)²+1-(3a-1)]/(3a-1) (将a²=3a-1 带入并化简)
=3x(18a-6)/(3a-1)
=3x6
=18
所以 a³/(a^6+1)=1/ (a^6+1)/a³
=1/18

Answer 2

a²-3a+1=0
∴a²=3a-1
a³=a(3a-1)=3a²-a=3（3a-1）-a=8a-3
a³/（a的六次方+1）
=（8a-3）/[(8a-3)²+1]
=(8a-3)/[64a²-48a+10]
=（8a-3）/(192a-48a-54)
=(8a-3)/(144a-54)
=(8a-3)/18(8a-3)
=1/18

Answer 3

解：b²－4ac＝（－3）²－4×1×1＝5
则√（b²－4ac）＝√5
则－b＋√（b²－4ac）＝3＋√5，－b－√（b²－4ac）＝3－√5
∴a1＝（3＋√5）/2，a2＝（3－√5）/2
当a＝（3＋√5）/2时，
原式＝[（3＋√5）/2]³÷{[（3＋√5）/2]^6＋1}＝（9＋4√5）÷[（162＋72√5）/16]＝（72＋32√5）/（81＋36√5）
当a＝（3－√5）/2时
原式＝[（3－√5）/2]³÷{[（3－√5）/2]^6＋1}＝（9－4√5）÷[（162－72√5）/16]＝（72－32√5）/（81＋36√5

Answer 4

a²-3a+1=0，
A-3+1/A=0 ==>A+1/A=3 ==>(A+1/A)^2=3 ==>A^2+2+1/A^2=9 ==>A^2+1/A^2=7

A^3/(A^6+1)=1/(A^3+1/A^3)=1/((A+1/A)(A^2-1+1/A^2)=1/(3*(7-1)=1/18

Answer 5

a^2=3a-1
a^6+1=(3a-1)^3+1
=(27a^3-1-27a^2+9a)+1
=27a^3-27a^2+9a
=9a(3a^2-3a+1)
=9a[3a^2-(3a-1)]
=9a*2a^2
a^3/(a^6+1)=a3/(9a*2a^2)=1/18