已知x2-x-1=0,则x4+2x+1/x5的值为?
2个回答
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x^2-x-1=0
(x-1/2)^2-5/4=0
x-1/2=1/2*根号5 或x-1/2=-1/2*根号5
x=1/2+1/2*根号5 或x=1/2-1/2*根号5
将x的值代入:
x^4+2x+1/x5
=(x+1)^2+2x+1/[x*(x+1)^2]
=x^2+2x+1+2x+1/(x^3+2x^2+x)
=x+1+4x+1+1/[x*(x+1)+2(x+1)+x]
=5x+2+1/(x+1+x+2x+2+x)
=5x+2+1/(5x+3)
=(25x^2+15x+10x+6+1)/(5x+3)
=(25x+25+25x+7)/(5x+3)
=(50x+32)/(5x+3)
=10+2/(5x+3)
(很复杂,过程不知道是否出错)
(x-1/2)^2-5/4=0
x-1/2=1/2*根号5 或x-1/2=-1/2*根号5
x=1/2+1/2*根号5 或x=1/2-1/2*根号5
将x的值代入:
x^4+2x+1/x5
=(x+1)^2+2x+1/[x*(x+1)^2]
=x^2+2x+1+2x+1/(x^3+2x^2+x)
=x+1+4x+1+1/[x*(x+1)+2(x+1)+x]
=5x+2+1/(x+1+x+2x+2+x)
=5x+2+1/(5x+3)
=(25x^2+15x+10x+6+1)/(5x+3)
=(25x+25+25x+7)/(5x+3)
=(50x+32)/(5x+3)
=10+2/(5x+3)
(很复杂,过程不知道是否出错)
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