(x^2+1/x^2)+(x+1/x)=4(x+1/x)^2-2+(x+1/x)=4令x+1/x=tt^2+t-6=0(t+3)(t-2)=0t=-3,t=2x+1/x=-3,x+1/x=2x^2+3x+1=0,x^2-2x+1=0x=(-3±√5)/2,x=1(二重根)
x^2+1/x^2+x+1/x=4(x^2+1/x^2+2)+(x+1/x)+1/4=4+2+1/4((x+1/x)+1/2)^2=(5/2)^2(x+1/x)+1/2=±5/2x=1;
x=1