1. Let the decimal part of $\frac{2}{\sqrt{5}-1}$ be denoted as $\alpha$. Then the value of $\alpha(\alpha+1)$ is ( ).
(A) 1
(B) a positive rational number less than 1
(C) a positive irrational number less than 1
(D) a positive irrational number greater than 1
-1.A.
From $\frac{2}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{2}=1+\frac{\sqrt{5}-1}{2}$, and $0<\frac{\sqrt{5}-1}{2}<\frac{3-1}{2}=1$,
we get $\alpha=\frac{\sqrt{5}-1}{2}=\frac{2}{\sqrt{5}+1}, \alpha+1=\frac{2}{\sqrt{5}-1}$.
Thus, $\alpha(\alpha+1)=\frac{2}{\sqrt{5}+1} \times \frac{2}{\sqrt{5}-1}=1$.