[Solution] $\because \alpha \beta+\alpha-\beta=\alpha \beta+\alpha-\beta-1+1=(\alpha-1)(\beta+1)+1$, by setting $\alpha=1+\sqrt{2}, \beta=-1+\sqrt{2}$, then $\alpha \beta+\alpha-\beta$ is a rational number, so (A) is incorrect;

If we set $\alpha=2 \sqrt{2}, \beta=\sqrt{2}$, then $\frac{\alpha-\beta}{\alpha+\beta}$ is a rational number. So (B) is incorrect; and if we set $\alpha=\sqrt[3]{2}, \beta=-\sqrt{2}$, then $\sqrt{\alpha}+\sqrt[3]{\beta}=0$ is a rational number. So (C) is incorrect. Therefore, the answer is (A).
$1 \cdot 61$ There are the following three propositions:
(甲) If $\alpha, \beta$ are unequal irrational numbers, then $\alpha \beta + \alpha - \beta$ is an irrational number;
(乙) If $\alpha, \beta$ are unequal irrational numbers, then $\frac{\alpha - \beta}{\alpha + \beta}$ is an irrational number;
(丙) If $\alpha, \beta$ are unequal irrational numbers, then $\sqrt{a} + \sqrt[3]{\beta}$ is an irrational number. The number of correct propositions is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(China Junior High School Mathematics League, 1999)