1. The three sides of a triangle are exactly $m^{2}+m+1, 2m+1, m^{2}-1$. Then the largest angle of this triangle is ( ).
A. $\frac{\pi}{3}$
B. $\frac{2 \pi}{3}$
C. $\frac{3 \pi}{4}$
1. $\frac{5 \pi}{6}$
1. B

Obviously, $m^{2}+m+1, 2m+1, m^{2}-1$ are all positive, so $m>1$.
It is easy to see that $m^{2}+m+1$ is the largest side of this triangle. Let the angle opposite to it be $\alpha$, then $\cos \alpha=\frac{\left(m^{2}-1\right)^{2}+(2 m+1)^{2}-\left(m^{2}+m+1\right)^{2}}{2\left(m^{2}-1\right)(2 m+1)}=-\frac{1}{2}$, so $\alpha=\frac{2 \pi}{3}$, choose B