若 $\cos (\alpha-\beta)=\frac{\sqrt{5}}{5}, \cos 2 \alpha=\frac{\sqrt{10}}{10}$, 且 $\alpha 、 \beta$ 均为锐角, $\alpha<\beta$, 则 $\alpha+\beta$ 的值为 ( C )
析] $\because 0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}, \quad \alpha<\beta$,

$\therefore-\frac{\pi}{2}<\alpha-\beta<0$.

又 $\cos (\alpha-\beta)=\frac{\sqrt{5}}{5}$,

$\therefore \sin (\alpha-\beta)=-\sqrt{1-\cos ^{2}(\alpha-\beta)}=-\frac{2 \sqrt{5}}{5}$.
又 $\because 0<2 \alpha<\pi, \cos 2 \alpha=\frac{\sqrt{10}}{10}$,

$\therefore \sin 2 \alpha=\sqrt{1-\cos ^{2} 2 \alpha}=\frac{3 \sqrt{10}}{10}$.

$\therefore \cos (\alpha+\beta)=\cos [2 \alpha-(\alpha-\beta)]=\cos 2 \alpha \cos (\alpha-\beta)+\sin 2 \alpha \sin (\alpha-\beta)=\frac{\sqrt{10}}{10} \times \frac{\sqrt{5}}{5}+\frac{3 \sqrt{10}}{10} \times\left(-\frac{2 \sqrt{5}}{5}\right)=-\frac{\sqrt{2}}{2}$.

又 $0<\alpha+\beta<\pi$, 故 $\alpha+\beta=\frac{3 \pi}{4}$.