解：∵x²-3x+1=0∴x²+1=3x(x²+1)×1/x=3x×1/xx²×1/x+1×1/x=3∴x+(1/x)=3∴[x+(1/x)]²=3²x²+2×x×1/x+(1/x²)=9x²+2+(1/x²)=9∴x²+(1/x²)=9-2=7
