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7. 魔术师在表演中请观众任意想一个数，然后将这个数按照以下步骤操作，魔术师立刻说出了观众想的那个数.
小乐想了一个数，并告诉魔术师结果为80，则小乐想的这个数是
小乐想了一个数，并告诉魔术师结果为80，则小乐想的这个数是
75
.答案
7. 75 解析:$[(80-7)×4+8]÷4=(292+8)÷4=75$.
8. 计算：
(1) $-2^{2} ÷ \dfrac{4}{7} × (\dfrac{3}{4} - \dfrac{4}{7}) + 3^{2} ÷ (-9)$；
(2) $-\dfrac{3}{4} × [-3^{2} × (-\dfrac{2}{3})^{2} - 2]$；
(3) $-\dfrac{1}{42} ÷ (\dfrac{1}{6} - \dfrac{3}{14} + \dfrac{2}{7})$；
(4) $[2\dfrac{1}{2} - (\dfrac{7}{9} - \dfrac{11}{12} + \dfrac{1}{6}) × 36] ÷ 5$。
(1) $-2^{2} ÷ \dfrac{4}{7} × (\dfrac{3}{4} - \dfrac{4}{7}) + 3^{2} ÷ (-9)$；
(2) $-\dfrac{3}{4} × [-3^{2} × (-\dfrac{2}{3})^{2} - 2]$；
(3) $-\dfrac{1}{42} ÷ (\dfrac{1}{6} - \dfrac{3}{14} + \dfrac{2}{7})$；
(4) $[2\dfrac{1}{2} - (\dfrac{7}{9} - \dfrac{11}{12} + \dfrac{1}{6}) × 36] ÷ 5$。
答案
8. (1)原式$=-4×\dfrac{7}{4} ×(\dfrac{3}{4}-\dfrac{4}{7}) +9÷(-9) =-7×\dfrac{3}{4}-(-7)×\dfrac{4}{7}+(-1) =-\dfrac{21}{4}+4-1=-\dfrac{9}{4}$.
(2)原式$=-\dfrac{3}{4} ×(-9×\dfrac{4}{9}-2) =-\dfrac{3}{4} ×(-4-2) =\dfrac{9}{2}$.
(3)原式$=-\dfrac{1}{42} ÷\dfrac{7-9+12}{42} =-\dfrac{1}{42} ÷\dfrac{10}{42} =-\dfrac{1}{42} ×\dfrac{42}{10} =-\dfrac{1}{10}$.
(4)原式$=[\dfrac{5}{2}-(\dfrac{7}{9} ×36-\dfrac{11}{12} ×36+\dfrac{1}{6} ×36)] ×\dfrac{1}{5} =[\dfrac{5}{2}-(28-33+6)] ×\dfrac{1}{5} =\dfrac{3}{2} ×\dfrac{1}{5} =\dfrac{3}{10}$.
(2)原式$=-\dfrac{3}{4} ×(-9×\dfrac{4}{9}-2) =-\dfrac{3}{4} ×(-4-2) =\dfrac{9}{2}$.
(3)原式$=-\dfrac{1}{42} ÷\dfrac{7-9+12}{42} =-\dfrac{1}{42} ÷\dfrac{10}{42} =-\dfrac{1}{42} ×\dfrac{42}{10} =-\dfrac{1}{10}$.
(4)原式$=[\dfrac{5}{2}-(\dfrac{7}{9} ×36-\dfrac{11}{12} ×36+\dfrac{1}{6} ×36)] ×\dfrac{1}{5} =[\dfrac{5}{2}-(28-33+6)] ×\dfrac{1}{5} =\dfrac{3}{2} ×\dfrac{1}{5} =\dfrac{3}{10}$.
9. 先观察下列各式,再解答问题.
$\frac{1}{2 × 3}=\frac{1}{2}-\frac{1}{3} ; \frac{1}{3 × 4}=\frac{1}{3}-\frac{1}{4} ; \frac{1}{4 × 5}=\frac{1}{4}-\frac{1}{5}.$
(1) ①根据上面各式的规律填空: $\frac{1}{5 × 6}=$
②$\frac{1}{1 × 2}+\frac{1}{2 × 3}+\frac{1}{3 × 4}+\dots+\frac{1}{n(n+1)}=$
(2)运用以上方法,求 $\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}+\frac{1}{112}+\frac{1}{144}$ 的值.
$\frac{1}{2 × 3}=\frac{1}{2}-\frac{1}{3} ; \frac{1}{3 × 4}=\frac{1}{3}-\frac{1}{4} ; \frac{1}{4 × 5}=\frac{1}{4}-\frac{1}{5}.$
(1) ①根据上面各式的规律填空: $\frac{1}{5 × 6}=$
$\dfrac{1}{5}-\dfrac{1}{6}$
;②$\frac{1}{1 × 2}+\frac{1}{2 × 3}+\frac{1}{3 × 4}+\dots+\frac{1}{n(n+1)}=$
$\dfrac{n}{n+1}$
($n$ 为整数,且 $n ≥ 1$).(2)运用以上方法,求 $\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}+\frac{1}{112}+\frac{1}{144}$ 的值.
答案
9. (1)①$\dfrac{1}{5}-\dfrac{1}{6}$ ②$\dfrac{n}{n+1}$ 解析:原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dots+\dfrac{1}{n}-\dfrac{1}{n+1}=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}$.
(2)原式$=\dfrac{1}{2} ×(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dots+\dfrac{1}{56}+\dfrac{1}{72}) =\dfrac{1}{2} ×(\dfrac{1}{1×2}+\dfrac{1}{2×3}+\dfrac{1}{3×4}+\dots+\dfrac{1}{7×8}+\dfrac{1}{8×9}) =\dfrac{1}{2} ×(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}) =\dfrac{1}{2} ×(1-\dfrac{1}{9}) =\dfrac{4}{9}$.
(2)原式$=\dfrac{1}{2} ×(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dots+\dfrac{1}{56}+\dfrac{1}{72}) =\dfrac{1}{2} ×(\dfrac{1}{1×2}+\dfrac{1}{2×3}+\dfrac{1}{3×4}+\dots+\dfrac{1}{7×8}+\dfrac{1}{8×9}) =\dfrac{1}{2} ×(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}) =\dfrac{1}{2} ×(1-\dfrac{1}{9}) =\dfrac{4}{9}$.
