已知x+x分之1=2 求x^3+x^3分之一
4个回答
展开全部
x^3 + 1/x^3
= (x + 1/x)^3 - 3 x^2/x - 3 x/x^2
= (x + 1/x)^3 - 3 (x + 3 /x)
= 2^3 - 3×2
=2
x^2 + 1/x^2
= (x + 1/x)^2 - 2 x 1/x
= 2^2 -2 = 2
x^n + 1/x^n
= (x^n + 1/x^n) (x + 1/x) / (x + 1/x)
= {x^(n+1) + 1/x^(n+1) + x^(n-1) + 1/x^(n-1)} / 2
得: x^(n+1) + 1/x^(n+1) = 2 {x^n + 1/x^n} - { x^(n-1) + 1/x^(n-1)}
由上式可知:
x^4 + 1/x^4 = 2 (x^3 + 1/x^3) - (x^2 + 1/x^2) = 2×2 - 2 = 2
x^5 + 1/x^5 = 2 (x^4 + 1/x^4) - (x^3 + 1/x^3) = 2×2 - 2 = 2
..........
x^n + 1/x^n = 2
= (x + 1/x)^3 - 3 x^2/x - 3 x/x^2
= (x + 1/x)^3 - 3 (x + 3 /x)
= 2^3 - 3×2
=2
x^2 + 1/x^2
= (x + 1/x)^2 - 2 x 1/x
= 2^2 -2 = 2
x^n + 1/x^n
= (x^n + 1/x^n) (x + 1/x) / (x + 1/x)
= {x^(n+1) + 1/x^(n+1) + x^(n-1) + 1/x^(n-1)} / 2
得: x^(n+1) + 1/x^(n+1) = 2 {x^n + 1/x^n} - { x^(n-1) + 1/x^(n-1)}
由上式可知:
x^4 + 1/x^4 = 2 (x^3 + 1/x^3) - (x^2 + 1/x^2) = 2×2 - 2 = 2
x^5 + 1/x^5 = 2 (x^4 + 1/x^4) - (x^3 + 1/x^3) = 2×2 - 2 = 2
..........
x^n + 1/x^n = 2
展开全部
x+1/x=2经过通分得(x²+1)/x=2, 移项得x²+1-2x=0,所以(x-1)²=0,所以x=1.
众所周知,1的多少次方都是1,所以结果可想而知
众所周知,1的多少次方都是1,所以结果可想而知
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
2011-04-22 · 知道合伙人教育行家
关注
展开全部
x+1/x = 2
(x+1/x)^2 = 2^2
x^2+1/x^2+2=4
x^2+1/x^2=2
x^3+1/x^3=(x+1/x)(x^2-x*1/x+1/x^2)
= (x+1/x) { (x^2+1/x^2)-1 }
= (x+1/x) { (x^2+1/x^2)-1 }
= 2/(2-1)
=2
(x+1/x)^2 = 2^2
x^2+1/x^2+2=4
x^2+1/x^2=2
x^3+1/x^3=(x+1/x)(x^2-x*1/x+1/x^2)
= (x+1/x) { (x^2+1/x^2)-1 }
= (x+1/x) { (x^2+1/x^2)-1 }
= 2/(2-1)
=2
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
(x+1/x)^3=2^3
x^3+3x +3/x+1/x^3=8
x^3+1/x^3+3(x +1/x)=8
x^3+1/x^3+3*2=8
x^3+1/x^3=8-6=2
x^3+3x +3/x+1/x^3=8
x^3+1/x^3+3(x +1/x)=8
x^3+1/x^3+3*2=8
x^3+1/x^3=8-6=2
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询