设y=x^2+2x+1/x^2-2x-1,
去分母,整理得(y-1)x²-(2y+2)x-(y+1)=0,
∵上述关于x的方程有解,
∴△=[-(2y+2)]²-4(y-1)[-(y+1)]=8y²+8y=8y(y+1)≥0,
∴y≤-1或y≥0,
故所求值域是(-∞,-1]∪[0,+∞).
