令t=√(x^2+2x+2p)>=0则：x^2+2x=t^2-2p代入原方程得：t^2+2t-p^2-2p=0得：(t+1)^2=(p+1)^2即t=p or t=-p-21）若方程无实根,则：√(x^2+2x+2p)=p,无实根,p=0,且x^2+2x+2p-p^2=0无实根,但此时(x+1)^2=(p-1)^2,得：x=p...