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已知X2-3x+1=0,求下列各式的值：（1）x2+1/x2;(2)x3+1/x3;(3)x5+1/x5
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已知X2-3x+1=0,求下列各式的值：（1）x2+1/x2;(2)x3+1/x3;(3)x5+1/x5
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答案和解析
X2-3x+1=0
x-3+1/x=0 两边除以x
∴x+1/x=3
x²+1/x²=(x+1/x)²-2=3²-2=7
x³+1/x³=(x+1/x)(x²-1+1/x²)=3×6=18
(x²+1/x²)(x³+1/x³)=x5+1/x+x+1/x5
∴7×18=x5+3+1/x5
∴x5+1/x5=123
x-3+1/x=0 两边除以x
∴x+1/x=3
x²+1/x²=(x+1/x)²-2=3²-2=7
x³+1/x³=(x+1/x)(x²-1+1/x²)=3×6=18
(x²+1/x²)(x³+1/x³)=x5+1/x+x+1/x5
∴7×18=x5+3+1/x5
∴x5+1/x5=123
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