已知根号X+根号x分之1=5,求x平方+x+1分之x-x+1分之x的值
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2010-10-31 · 知道合伙人教育行家
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根号X+1/根号x=5,两边平方得:
x+1/x+2=25
x+1/x=23
x^2+1/x^2+2=529
x^2+1/x^2=527
x/(x^2+x+1)-x/(x^2-x+1)
=1/(x+1+1/x)-1/(x-1+1/x)
=1/[(x+1/x)+1]-1/[(x+1/x)-1]
=1/(23+1)-1/(23-1)
=1/24-1/22
=-1/264
【问题补充:如果已知条件不变,求根号x的平方+x+1分之x 减 根号x的平方-x+1分之x的值】
根号[x/(x^2+x+1)]-根号[x/(x^2-x+1)]
=根号[1/(x+1+1/x)]-根号[1/(x-1+1/x)]
=根号[1/(x+1/x+1)]-根号[1/(x+1/x-1)]
=根号(1/(23+1)]-根号[1/(23-1)]
=1/根号24-1/根号22
=(根号22-2根号6)/(4根号33)
x+1/x+2=25
x+1/x=23
x^2+1/x^2+2=529
x^2+1/x^2=527
x/(x^2+x+1)-x/(x^2-x+1)
=1/(x+1+1/x)-1/(x-1+1/x)
=1/[(x+1/x)+1]-1/[(x+1/x)-1]
=1/(23+1)-1/(23-1)
=1/24-1/22
=-1/264
【问题补充:如果已知条件不变,求根号x的平方+x+1分之x 减 根号x的平方-x+1分之x的值】
根号[x/(x^2+x+1)]-根号[x/(x^2-x+1)]
=根号[1/(x+1+1/x)]-根号[1/(x-1+1/x)]
=根号[1/(x+1/x+1)]-根号[1/(x+1/x-1)]
=根号(1/(23+1)]-根号[1/(23-1)]
=1/根号24-1/根号22
=(根号22-2根号6)/(4根号33)
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