$2 \cdot 107$ If the three roots of the equation $(x-1)\left(x^{2}-2 x+m\right)=0$ can be the lengths of the three sides of a triangle, then the range of the real number $m$ is
(A) $0 \leqslant m \leqslant 1$.
(B) $m \geqslant \frac{3}{4}$.
(C) $\frac{3}{4}<m \leqslant 1$.
(D) $\frac{3}{4} \leqslant m \leqslant 1$.
(China Junior High School Mathematics League, 1995)
[Solution 1] Let $\alpha, \beta$ be the roots of the equation $x^{2}-2 x+m=0$. From $\Delta=4-4 m \geqslant 0$, we get $m \leqslant 1$.
Obviously, $\alpha+\beta=2>1$,
and by the triangle inequality, which states that the difference of any two sides is less than the third side, we know $|\alpha-\beta|<1$. 
Thus, $m>\frac{3}{4}$.
Therefore, the answer is $(C)$.
[Solution 2] $\because x^{2}-2 x+m=0$ has two real roots, then $\Delta=4-m \geqslant 0$, we get $m \leqslant 1$.
The roots of the original equation are
$$
x_{1}=1, x_{2}=1-\sqrt{1-m}, x_{3}=1+\sqrt{1-m} \text {, clearly } x_{2} \leqslant
$$
$x_{1} \leqslant x_{3}$.
Noting that $x_{1}+x_{3}=2-\sqrt{1-m}>1+\sqrt{1-m}=x_{3}$,
we thus get $m>\frac{3}{4}$.
Therefore, the answer is $(C)$.