已知x(n次方)+1/x(2n次方)=1, 求x(5n次方)+x(n次方)+1
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x(5n次方)+x(n次方)+1=0
由x(n次方)+1/x(2n次方)=1可得
x(n次方)=1-1/x(2n次方);x(3n次方)+1=x(2n次方);x(5n次方)+x(2n次方)=x(4n次方)
所以x(5n次方)+x(n次方)+1=x(4n次方)-x(2n次方)+x(n次方)+x(2n次方)-x(3n次方)=x(3n次方)×[x(n次方)-1]+x(n次方)=x(n次方)-x(n次方)=0
由x(n次方)+1/x(2n次方)=1可得
x(n次方)=1-1/x(2n次方);x(3n次方)+1=x(2n次方);x(5n次方)+x(2n次方)=x(4n次方)
所以x(5n次方)+x(n次方)+1=x(4n次方)-x(2n次方)+x(n次方)+x(2n次方)-x(3n次方)=x(3n次方)×[x(n次方)-1]+x(n次方)=x(n次方)-x(n次方)=0
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