若x^2-7x+1=0,则x^4+1/x^4的个位数字是?

伱脸鎭荭：x²-7x+1=0(x≠0)两边同时除以x,得x-7+(1/x)=0x+(1/x)=7∴x²+(1/x²)=[x+(1/x)]²-2=7²-2=47∴x⁴+(1/x⁴)=[x²+(1/x²)]-2=47²-2=2207∴x⁴+(1/xS...