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4. 计算：
(1) $2\dfrac{1}{3} × \dfrac{7}{13} - (-\dfrac{7}{13}) × (-14) + (-\dfrac{4}{3}) ÷ \dfrac{13}{7}$。
(2) $0.7 × 19\dfrac{5}{9} + 2\dfrac{3}{4} × (-14) + \dfrac{7}{10} × \dfrac{4}{9} - 3.25 × 14$。
(3) $(-99\dfrac{22}{23}) × (-69) + 1.25 × (-\dfrac{81}{20}) × (-8)$。
(4) $(-\dfrac{1}{84}) ÷ (\dfrac{2}{3} + $
$ - \dfrac{5}{28} - \dfrac{1}{4})$。
(1) $2\dfrac{1}{3} × \dfrac{7}{13} - (-\dfrac{7}{13}) × (-14) + (-\dfrac{4}{3}) ÷ \dfrac{13}{7}$。
(2) $0.7 × 19\dfrac{5}{9} + 2\dfrac{3}{4} × (-14) + \dfrac{7}{10} × \dfrac{4}{9} - 3.25 × 14$。
(3) $(-99\dfrac{22}{23}) × (-69) + 1.25 × (-\dfrac{81}{20}) × (-8)$。
(4) $(-\dfrac{1}{84}) ÷ (\dfrac{2}{3} + $
$ - \dfrac{5}{28} - \dfrac{1}{4})$。答案
(1) 原式$=\dfrac{7}{3}× \dfrac{7}{13}-\dfrac{7}{13}× 14-\dfrac{4}{3}× \dfrac{7}{13}=\dfrac{7}{13}× (\dfrac{7}{3}-14-\dfrac{4}{3})=\dfrac{7}{13}× (-13)=-7$.
(2) 原式$=0.7× (19\dfrac{5}{9}+\dfrac{4}{9})-14× (2\dfrac{3}{4}+3\dfrac{1}{4})=0.7× 20-14× 6=14-84=-70$.
(3) 原式$=(100-\dfrac{1}{23})× 69-1.25× 8× (-\dfrac{81}{20})=6900-3-10× (-\dfrac{81}{20})=6900-3+\dfrac{81}{2}=6937.5$.
(4) 因为原式的倒数为$(\dfrac{2}{3}+\dfrac{3}{7}-\dfrac{5}{28}-\dfrac{1}{4})÷ (-\dfrac{1}{84})=(\dfrac{2}{3}+\dfrac{3}{7}-\dfrac{5}{28}-\dfrac{1}{4})× (-84)=\dfrac{2}{3}× (-84)+\dfrac{3}{7}× (-84)-\dfrac{5}{28}× (-84)-\dfrac{1}{4}× (-84)=-56-36+15+21=-56$,所以原式$=-\dfrac{1}{56}$.
(2) 原式$=0.7× (19\dfrac{5}{9}+\dfrac{4}{9})-14× (2\dfrac{3}{4}+3\dfrac{1}{4})=0.7× 20-14× 6=14-84=-70$.
(3) 原式$=(100-\dfrac{1}{23})× 69-1.25× 8× (-\dfrac{81}{20})=6900-3-10× (-\dfrac{81}{20})=6900-3+\dfrac{81}{2}=6937.5$.
(4) 因为原式的倒数为$(\dfrac{2}{3}+\dfrac{3}{7}-\dfrac{5}{28}-\dfrac{1}{4})÷ (-\dfrac{1}{84})=(\dfrac{2}{3}+\dfrac{3}{7}-\dfrac{5}{28}-\dfrac{1}{4})× (-84)=\dfrac{2}{3}× (-84)+\dfrac{3}{7}× (-84)-\dfrac{5}{28}× (-84)-\dfrac{1}{4}× (-84)=-56-36+15+21=-56$,所以原式$=-\dfrac{1}{56}$.
5. 观察下列各式：
$\frac{1}{1×2}=1-\frac{1}{2}$，$\frac{1}{2×3}=\frac{1}{2}-\frac{1}{3}$，$\frac{1}{3×4}=\frac{1}{3}-\frac{1}{4}$，$\frac{1}{4×5}=\frac{1}{4}-\frac{1}{5}$，…。
探索规律，根据规律解答下列问题：
（1）第6个等式为
（2）计算：$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{2025×2026}$。
（3）若有理数$a,b$满足$|a-3|+|b-5|=0$，求$\frac{1}{ab}+\frac{1}{(a+2)(b+2)}+\frac{1}{(a+4)(b+4)}+…+$
$$的值。
$\frac{1}{1×2}=1-\frac{1}{2}$，$\frac{1}{2×3}=\frac{1}{2}-\frac{1}{3}$，$\frac{1}{3×4}=\frac{1}{3}-\frac{1}{4}$，$\frac{1}{4×5}=\frac{1}{4}-\frac{1}{5}$，…。
探索规律，根据规律解答下列问题：
（1）第6个等式为
$\frac{1}{6×7}$
=$\frac{1}{6}-\frac{1}{7}$
。（2）计算：$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{2025×2026}$。
（3）若有理数$a,b$满足$|a-3|+|b-5|=0$，求$\frac{1}{ab}+\frac{1}{(a+2)(b+2)}+\frac{1}{(a+4)(b+4)}+…+$
$$的值。答案
(1) $\dfrac{1}{6× 7}$；$\dfrac{1}{6}-\dfrac{1}{7}$.
(2) 原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots +\dfrac{1}{2025}-\dfrac{1}{2026}=1-\dfrac{1}{2026}=\dfrac{2025}{2026}$.
(3) 因为$|a-3|+|b-5|=0$,所以$a-3=0,b-5=0$. 所以$a=3,b=5$. 所以原式$=\dfrac{1}{3× 5}+\dfrac{1}{5× 7}+\dfrac{1}{7× 9}+\dots +\dfrac{1}{103× 105}=\dfrac{1}{2}× (\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dots +\dfrac{1}{103}-\dfrac{1}{105})=\dfrac{1}{2}× (\dfrac{1}{3}-\dfrac{1}{105})=\dfrac{1}{2}× \dfrac{34}{105}=\dfrac{17}{105}$.
(2) 原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots +\dfrac{1}{2025}-\dfrac{1}{2026}=1-\dfrac{1}{2026}=\dfrac{2025}{2026}$.
(3) 因为$|a-3|+|b-5|=0$,所以$a-3=0,b-5=0$. 所以$a=3,b=5$. 所以原式$=\dfrac{1}{3× 5}+\dfrac{1}{5× 7}+\dfrac{1}{7× 9}+\dots +\dfrac{1}{103× 105}=\dfrac{1}{2}× (\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dots +\dfrac{1}{103}-\dfrac{1}{105})=\dfrac{1}{2}× (\dfrac{1}{3}-\dfrac{1}{105})=\dfrac{1}{2}× \dfrac{34}{105}=\dfrac{17}{105}$.
