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三、解答题（共40分）
13. (8分)将下列各数填在相应的括号内.
$-3.8,-20\%,4.3,-\left\lvert -\dfrac{20}{7}\right\rvert,4^2,0,-(-\dfrac{3}{5}),-3^2,\dfrac{π}{2}.$
非负整数:{ };
负分数:{ };
正有理数:{ };
负有理数:{ }.
13. (8分)将下列各数填在相应的括号内.
$-3.8,-20\%,4.3,-\left\lvert -\dfrac{20}{7}\right\rvert,4^2,0,-(-\dfrac{3}{5}),-3^2,\dfrac{π}{2}.$
非负整数:{ };
负分数:{ };
正有理数:{ };
负有理数:{ }.
答案
13.非负整数：$4^2,0$
负分数：$-3.8,-20\%,-\left|-\dfrac{20}{7}\right|$
正有理数：$4.3,4^2,-(-\dfrac{3}{5})$
负有理数：$-3.8,-20\%,-\left|-\dfrac{20}{7}\right|,-3^2$
负分数：$-3.8,-20\%,-\left|-\dfrac{20}{7}\right|$
正有理数：$4.3,4^2,-(-\dfrac{3}{5})$
负有理数：$-3.8,-20\%,-\left|-\dfrac{20}{7}\right|,-3^2$
14. (20 分)计算：
(1)$8 - (-3) + (-2)$；
(2)$-10 + 8 ÷ (-2)^2 - 4 × 3$；
(3)$(\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{6}) ÷ (-\dfrac{1}{18})$；
(4)$-10^2 + [(-4)^2 - (1 - 3^2) ÷ \dfrac{1}{2}].$
(1)$8 - (-3) + (-2)$；
(2)$-10 + 8 ÷ (-2)^2 - 4 × 3$；
(3)$(\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{6}) ÷ (-\dfrac{1}{18})$；
(4)$-10^2 + [(-4)^2 - (1 - 3^2) ÷ \dfrac{1}{2}].$
答案
14.(1)原式$=8+3-2=9$.
(2)原式$=-10+8÷4-12=-10+2-12=-20$.
(3)原式$=(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6})×(-18)=\dfrac{1}{2}×(-18)+\dfrac{1}{3}×(-18)-\dfrac{1}{6}×(-18)=-9-6+3=-12$.
(4)原式$=-100+[16-(1-9)×2]=-100+16+16=-68$.
(2)原式$=-10+8÷4-12=-10+2-12=-20$.
(3)原式$=(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6})×(-18)=\dfrac{1}{2}×(-18)+\dfrac{1}{3}×(-18)-\dfrac{1}{6}×(-18)=-9-6+3=-12$.
(4)原式$=-100+[16-(1-9)×2]=-100+16+16=-68$.
15. (12 分)观察下列等式:$\dfrac{1}{1× 2}=1-\dfrac{1}{2}$,$\dfrac{1}{2× 3}=\dfrac{1}{2}-\dfrac{1}{3}$,$\dfrac{1}{3× 4}=\dfrac{1}{3}-\dfrac{1}{4}$,将以上三个等式两边分别相加,得$\dfrac{1}{1× 2}+\dfrac{1}{2× 3}+\dfrac{1}{3× 4}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}=1-\dfrac{1}{4}=\dfrac{3}{4}$.
(1)猜想并写出$\dfrac{1}{n(n+1)}=$
(2)直接写出下列各式的计算结果:
①$\dfrac{1}{1× 2}+\dfrac{1}{2× 3}+\dfrac{1}{3× 4}+\dots+\dfrac{1}{2\,026× 2\,027}=$
②$\dfrac{1}{1× 2}+\dfrac{1}{2× 3}+\dfrac{1}{3× 4}+\dots+\dfrac{1}{n(n+1)}=$
(3)探究并计算:$\dfrac{1}{2× 4}+\dfrac{1}{4× 6}+\dfrac{1}{6× 8}+\dots+\dfrac{1}{2\,026× 2\,028}$.
(1)猜想并写出$\dfrac{1}{n(n+1)}=$
$\dfrac{1}{n}-\dfrac{1}{n+1}$
.(2)直接写出下列各式的计算结果:
①$\dfrac{1}{1× 2}+\dfrac{1}{2× 3}+\dfrac{1}{3× 4}+\dots+\dfrac{1}{2\,026× 2\,027}=$
$\dfrac{2026}{2027}$
;②$\dfrac{1}{1× 2}+\dfrac{1}{2× 3}+\dfrac{1}{3× 4}+\dots+\dfrac{1}{n(n+1)}=$
$\dfrac{n}{n+1}$
.(3)探究并计算:$\dfrac{1}{2× 4}+\dfrac{1}{4× 6}+\dfrac{1}{6× 8}+\dots+\dfrac{1}{2\,026× 2\,028}$.
答案
15.(1)$\dfrac{1}{n}-\dfrac{1}{n+1}$
(2)①$\dfrac{2026}{2027}$ 解析：原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{2026}-\dfrac{1}{2027}=1-\dfrac{1}{2027}=\dfrac{2026}{2027}$.
②$\dfrac{n}{n+1}$ 解析：原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{n}-\dfrac{1}{n+1}=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}$.
(3)原式$=\dfrac{1}{2}×(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dots+\dfrac{1}{2026}-\dfrac{1}{2028})=\dfrac{1}{2}×(\dfrac{1}{2}-\dfrac{1}{2028})=\dfrac{1}{2}×\dfrac{1013}{2028}=\dfrac{1013}{4056}$.
(2)①$\dfrac{2026}{2027}$ 解析：原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{2026}-\dfrac{1}{2027}=1-\dfrac{1}{2027}=\dfrac{2026}{2027}$.
②$\dfrac{n}{n+1}$ 解析：原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{n}-\dfrac{1}{n+1}=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}$.
(3)原式$=\dfrac{1}{2}×(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dots+\dfrac{1}{2026}-\dfrac{1}{2028})=\dfrac{1}{2}×(\dfrac{1}{2}-\dfrac{1}{2028})=\dfrac{1}{2}×\dfrac{1013}{2028}=\dfrac{1013}{4056}$.
