《2025年综合应用创新题典中点六年级数学上册鲁教版五四制》

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

答案： 【解】
(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：$50÷(\frac {1}{3}-\frac {1}{4}+\frac {1}{12})$.
解法 1：原式$=50÷\frac {1}{3}-50÷\frac {1}{4}+50÷\frac {1}{12}= 50×3-50×4+50×12= 550$.
解法 2：原式的倒数$=(\frac {1}{3}-\frac {1}{4}+\frac {1}{12})÷50= \frac {1}{3}×\frac {1}{50}-\frac {1}{4}×\frac {1}{50}+\frac {1}{12}×\frac {1}{50}= \frac {1}{300}$,故原式$=300$.
(1)以上两种解法中,解法2是正确的；
(2)请你计算：$(-\frac {7}{8})÷(1\frac {3}{4}-\frac {7}{8}-\frac {7}{12})$.
【解】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.

答案： 【解】
(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})+... +(\frac {1}{60}+\frac {2}{60}+... +\frac {58}{60}+\frac {59}{60})$.

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

11. (1)$\frac {1}{2}×\frac {2}{3}=$$\frac{1}{3}$；
$\frac {1}{2}×\frac {2}{3}×\frac {3}{4}=$$\frac{1}{4}$；
$\frac {1}{2}×\frac {2}{3}×\frac {3}{4}×\frac {4}{5}=$$\frac{1}{5}$；
猜想：$\frac {1}{2}×\frac {2}{3}×\frac {3}{4}×\frac {4}{5}×... ×\frac {n}{n+1}=$$\frac{1}{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)×... ×$$(\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？
(2)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.

答案： 【解】
(1)$\frac{1}{3}$;$\frac{1}{4}$;$\frac{1}{5}$;$\frac{1}{n+1}$
(2)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.