a=4的三次方根+2的三次方根+1的三次方根，求3/a + 3/a2 + 3/a3

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解决时间 2021-02-23 02:22

提问者网友：树红树绿
2021-02-22 10:15

a=4的三次方根+2的三次方根+1的三次方根，求3/a + 3/a*2 + 3/a*3

最佳答案

五星知识达人网友：不如潦草
2021-02-22 10:30

a+3/a) ②
把①代入②,得
3/[2^(2/(2-1)
=11[2^(1/3)-1]/2a
=(11/2)×{1/3)-1]}
=(11/2)×[2^(1/解:a=4的三次方根+2的三次方根+1的三次方根
=(2^2)^(1/3)+2^(1/3)+1
=2^(2/3)+2^(1/3)+1 ①
∴3/a + 3/a*2 + 3/a*3=3/a + 3/a*2 + 1/a
=4/2)×(1/3)-1]/3)-1]/a*3
=(11/3)+2^(1/a + 3/{[2^(2/3)+2^(1/3)+1][2^(1/2a=11/a*2 + 3/3)+1]}
=(11/2)×[2^(1/

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1楼网友：西岸风
2021-02-22 11:28

2/a + 2/a*2 + 1/a*3是不是写错了？a*2、a*3意思是a的2次方和3次方吧 a=4的三次方根+2的三次方根+1的三次方根，即a=4^(1/3)+2^(1/3) +1^(1/3) =2^(2/3) +2^(1/3) +1 由立方差公式可以知道x^3 -y^3=(x-y)*(x^2 +xy +y^2)，即 x^2 +xy +y^2 = (x^3 -y^3)/(x-y) 所以 a=2^(2/3) +2^(1/3) +1 = { [2^(1/3)]^3 -1} / [2^(1/3) -1] = 1/ [2^(1/3) -1] 故 2/a + 2/a^2 + 1/a^3 =2*[2^(1/3) -1] + 2*[2^(1/3) -1]^2 + [2^(1/3) -1]^3 =2*2^(1/3) -2 + 2*2^(2/3) -4*2^(1/3) +2 + 2 - 3 *2^(2/3) +3 *2^(1/3) -1 = -2^(2/3) + 2^(1/3) +1

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