8. 计算：
(1)$-29× 588+28× 588$；
(2)$-2\,026× \frac{3}{7}+2\,026× (-\frac{6}{7})+2\,026× \frac{2}{7}$。

(1)$-29×588+28×588$
$=588×(-29+28)$
$=588×(-1)$
$=-588.$
(2)$-2026×\frac{3}{7}+2026×(-\frac{6}{7})+2026×\frac{2}{7}$
$=2026×(-\frac{3}{7}-\frac{6}{7}+\frac{2}{7})$
$=2026×(-1)$
$=-2026.$

9. 阅读下列材料.
计算: $50 ÷ ( \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{12} )$.
解法1: 原式 $=50 ÷ \dfrac{1}{3} -50 ÷ \dfrac{1}{4} +50 ÷ \dfrac{1}{12}=50 × 3 -50 × 4 +50 × 12=550$.
解法2: 原式的倒数 $=( \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{12} ) ÷ 50=\dfrac{1}{3} × \dfrac{1}{50} - \dfrac{1}{4} × \dfrac{1}{50} + \dfrac{1}{12} × \dfrac{1}{50}=\dfrac{1}{300}$, 故原式 $=300$.
(1) 以上两种解法中, 解法 是正确的;
(2) 请你计算: $( -\dfrac{7}{8} ) ÷ ( 1\dfrac{3}{4} - \dfrac{7}{8} - \dfrac{7}{12} )$.

(1)2
(2)原式的倒数$=(1\frac{3}{4}-\frac{7}{8}-\frac{7}{12})÷(-\frac{7}{8})=(\frac{7}{4}-\frac{7}{8}-\frac{7}{12})×(-\frac{8}{7})=\frac{7}{4}×(-\frac{8}{7})-\frac{7}{8}×(-\frac{8}{7})-\frac{7}{12}×(-\frac{8}{7})=-2+1+\frac{2}{3}=-\frac{1}{3}$,故原式$=-3.$

10. 计算:$\frac{1}{2}+(\frac{1}{3}+\frac{2}{3})+(\frac{1}{4}+\frac{2}{4}+\frac{3}{4})+(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5})+\dots+(\frac{1}{60}+\frac{2}{60}+\dots+\frac{58}{60}+\frac{59}{60}).$

将原式记为①式,把①式括号内的数倒序后,得
$\frac{1}{2}+(\frac{2}{3}+\frac{1}{3})+(\frac{3}{4}+\frac{2}{4}+\frac{1}{4})+(\frac{4}{5}+\frac{3}{5}+\frac{2}{5}+\frac{1}{5})+\dots+(\frac{59}{60}+\frac{58}{60}+\dots+\frac{2}{60}+\frac{1}{60}).$②
①+②,得$1+2+3+4+\dots+59=1770.$
所以$\frac{1}{2}+(\frac{1}{3}+\frac{2}{3})+(\frac{1}{4}+\frac{2}{4}+\frac{3}{4})+(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5})+\dots+(\frac{1}{60}+\frac{2}{60}+\dots+\frac{58}{60}+\frac{59}{60})=\frac{1}{2}×1770=885.$

11. (1) $\frac{1}{2} × \frac{2}{3} =$ ；
$\frac{1}{2} × \frac{2}{3} × \frac{3}{4} =$ ；
$\frac{1}{2} × \frac{2}{3} × \frac{3}{4} × \frac{4}{5} =$ ；
猜想：$\frac{1}{2} × \frac{2}{3} × \frac{3}{4} × \frac{4}{5} × \dots × \frac{n}{n+1} =$ .
(2)根据上面的规律，解答下列问题：
①计算：$( \frac{1}{100} -1 ) × ( \frac{1}{99} -1 ) × ( \frac{1}{98} -1 ) × \dots × ( \frac{1}{4} -1 ) × ( \frac{1}{3} -1 ) × ( \frac{1}{2} -1 )$.
②将2026减去它的$\frac{1}{2}$，再减去余下的$\frac{1}{3}$，再减去余下的$\frac{1}{4}$，…，依次类推，最后减去余下的$\frac{1}{2026}$，则剩余的结果是多少？

(1)$\frac{1}{3}$；$\frac{1}{4}$；$\frac{1}{5}$；$\frac{1}{n+1}$
①$(\frac{1}{100}-1)×(\frac{1}{99}-1)×(\frac{1}{98}-1)×…×(\frac{1}{4}-1)×(\frac{1}{3}-1)×(\frac{1}{2}-1)$
$=-\frac{99}{100}×\frac{98}{99}×\frac{97}{98}×…×\frac{3}{4}×\frac{2}{3}×\frac{1}{2}$
$=-\frac{1}{100}.$
②依题意有$2026×(1-\frac{1}{2})×(1-\frac{1}{3})×…×(1-\frac{1}{2026})=2026×\frac{1}{2}×\frac{2}{3}×…×\frac{2025}{2026}=1.$