x²-3x+1=0,显然X是不等于0的,两边同时除以X,得x-3+(1/x)=0x+(1/x)=3两边同时平方,得x²+2+(1/x²)=9x²+(1/x²)=9-2=7
解:x²-3x+1=0 x-3+1/x=0 x+1/x=3 x²+1/x²=(x+1/x)²-2=7
等式两边同除以xx-3+1/x=0x+1/x=3x²+1/x²=x²+2x(1/x)+x²-2x(1/x)=(x+1/x)²-2=3²-2=7