10.计算:$1-2+3-4+5-6+\dots+2023-2024+2025.$

10. 原式$=(1-2)+(3-4)+(5-6)+\dots+(2\ 023-2\ 024)+2\ 025=(-1)+(-1)+(-1)+\dots+(-1)+2\ 025=-1\ 012+2\ 025=1\ 013.$

11. 若$\left| \frac{1}{2} -1 \right| = 1 - \frac{1}{2}$，$\left| \frac{1}{3} - \frac{1}{2} \right| = \frac{1}{2} - \frac{1}{3}$，$\left| \frac{1}{4} - \frac{1}{3} \right| = \frac{1}{3} - \frac{1}{4}$，…，照此规律回答下列问题：
问题：
(1) $\left| \frac{1}{19} - \frac{1}{18} \right| = \underline{\hspace{8cm}}$；
(2) 计算：$\left| \frac{1}{2} -1 \right| + \left| \frac{1}{3} - \frac{1}{2} \right| + \left| \frac{1}{4} - \frac{1}{3} \right| + \left| \frac{1}{5} - \frac{1}{4} \right|$；
(3) 计算：$\left| \frac{1}{2} -1 \right| + \left| \frac{1}{3} - \frac{1}{2} \right| + \left| \frac{1}{4} - \frac{1}{3} \right| + \dots + \left| \frac{1}{2023} - \frac{1}{2022} \right|$。

11. (1)$\dfrac{1}{18}-\dfrac{1}{19}$
(2) 原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}=1-\dfrac{1}{5}=\dfrac{4}{5}.$
(3) 原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{2\ 022}-\dfrac{1}{2\ 023}=1-\dfrac{1}{2\ 023}=\dfrac{2\ 022}{2\ 023}.$

12. 计算：$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\frac{1}{4×5}+…+\frac{1}{2023×2024}$

12. 原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dots+\dfrac{1}{2\ 023}-\dfrac{1}{2\ 024}=1-\dfrac{1}{2\ 024}=\dfrac{2\ 023}{2\ 024}.$

13. 计算：$\frac{1}{2} - \frac{1}{6} - \frac{1}{12} - \frac{1}{20} - \frac{1}{30}$

13. 原式$=\dfrac{1}{2}-\dfrac{1}{2×3}-\dfrac{1}{3×4}-\dfrac{1}{4×5}-\dfrac{1}{5×6}=\dfrac{1}{2}-(\dfrac{1}{2}-\dfrac{1}{3})-(\dfrac{1}{3}-\dfrac{1}{4})-(\dfrac{1}{4}-\dfrac{1}{5})-(\dfrac{1}{5}-\dfrac{1}{6})=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{1}{6}.$