2・13 If the equation $(m+2) x^{2}-2(m+1) x+m=0$ has only one real root, then the equation $(m+1) x^{2}-2 m x+m-2=0$
(A) has no real roots.
(B) has two distinct real roots.
(C) has two equal real roots.
(D) the number of real roots cannot be determined.
(China "Zu Chongzhi Cup" Junior High School Mathematics Invitational Competition, 1995)
[Solution] The discriminant of the original equation is $\Delta=4(m+1)^{2}-4 m(m+2)=4>0$, the equation has only one root, so the coefficient of the quadratic term is 0, i.e., $m+2=0$, giving $m=-2$.

Then the equation $(m+1) x^{2}-2 m x+m-2=0$ becomes $-x^{2}+4 x-4=0$ or $x^{2}-4 x+4=0$, which shows that the equation has two equal real roots.
Therefore, the correct choice is $(C)$.