《2025年经纶学典课时作业七年级数学上册苏科版》

4. 请用简便方法计算：
$\dfrac{1}{60}÷\left(\dfrac{2}{3}+\dfrac{4}{5}-\dfrac{7}{20}-\dfrac{5}{12}\right)$。

答案： $\frac{1}{42}$

5. 阅读材料：
求$1+2+2^{2}+2^{3}+2^{4}+…+2^{99}$的值。
解：设$S= 1+2+2^{2}+2^{3}+2^{4}+…+2^{99}$ ①,
将等式两边同时乘2,得
$2S= 2+2^{2}+2^{3}+2^{4}+…+2^{99}+2^{100}$ ②,
用②式减去①式,得$2S-S= 2^{100}-1$,
即$S= 1+2+2^{2}+2^{3}+2^{4}+…+2^{99}= 2^{100}-1$。
请你仿照此法计算：$1+3+3^{2}+3^{3}+3^{4}+…+3^{100}$。

答案： 设$S=1+3+3^{2}+3^{3}+3^{4}+\cdots +3^{100}$ ①，
将等式两边同时乘3,得
$3S=3+3^{2}+3^{3}+3^{4}+\cdots +3^{100}+3^{101}$ ②，
用②式减去①式,得$3S-S=3^{101}-1$,
即$S=1+3+3^{2}+3^{3}+3^{4}+\cdots +3^{100}=\frac{3^{101}-1}{2}$.

6. 请用简便方法计算：
(1)$\left(-1\dfrac{1}{2}\right)+\left(-2000\dfrac{5}{6}\right)+4000\dfrac{3}{4}+\left(-1999\dfrac{2}{3}\right)$；
(2)$99\dfrac{7}{8}×(-4)$。

答案：
(1)$\left(-1\frac{1}{2}\right)+\left(-2000\frac{5}{6}\right)+4000\frac{3}{4}+\left(-1999\frac{2}{3}\right)$
$=\left[-1+\left(-\frac{1}{2}\right)\right]+\left[-2000+\left(-\frac{5}{6}\right)\right]+\left(4000+\frac{3}{4}\right)+$
$\left[-1999+\left(-\frac{2}{3}\right)\right]$
$=[-1+(-2000)+4000+(-1999)]+\left[\left(-\frac{1}{2}\right)+\right.$
$\left(-\frac{5}{6}\right)+\frac{3}{4}+\left(-\frac{2}{3}\right)\right]$
$=0+\left(-\frac{5}{4}\right)=-\frac{5}{4}$.
(2)$99\frac{7}{8}× (-4)=\left(100-\frac{1}{8}\right)× (-4)=-399\frac{1}{2}$.

7. (1)观察并补全下面的算式：
$\dfrac{1}{2}-\dfrac{1}{3}=$$\frac{1}{6}$；$\dfrac{1}{3}-\dfrac{1}{4}=$$\frac{1}{12}$；
$\dfrac{1}{9}-\dfrac{1}{10}=$$\frac{1}{90}$；
$\dfrac{1}{2×3}=$$\frac{1}{6}$；$\dfrac{1}{3×4}=$$\frac{1}{12}$；
$\dfrac{1}{9×10}=$$\frac{1}{90}$。
(2)算式$\dfrac{1}{1×2}+\dfrac{1}{2×3}+\dfrac{1}{3×4}+…+\dfrac{1}{9×10}=$$\frac{9}{10}$。
(3)计算：$\dfrac{1}{3×7}+\dfrac{1}{7×11}+\dfrac{1}{11×15}+…+\dfrac{1}{55×59}$。
$\frac{1}{3× 7}+\frac{1}{7× 11}+\frac{1}{11× 15}+\cdots +\frac{1}{55× 59}$
$=\frac{1}{4}× \left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{4}× \left(\frac{1}{7}-\frac{1}{11}\right)+\frac{1}{4}× \left(\frac{1}{11}-\frac{1}{15}\right)+\cdots +\frac{1}{4}× \left(\frac{1}{55}-\frac{1}{59}\right)$
$=\frac{1}{4}× \left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\cdots +\frac{1}{55}-\frac{1}{59}\right)$
$=\frac{1}{4}× \left(\frac{1}{3}-\frac{1}{59}\right)=\frac{14}{177}$.
(4)计算：$-\dfrac{1}{3}-\dfrac{1}{15}-\dfrac{1}{35}-\dfrac{1}{63}-\dfrac{1}{99}$。
$-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}$
$=-\frac{1}{1× 3}-\frac{1}{3× 5}-\frac{1}{5× 7}-\frac{1}{7× 9}-\frac{1}{9× 11}$
$=-\frac{1}{2}× \left(1-\frac{1}{3}\right)-\frac{1}{2}× \left(\frac{1}{3}-\frac{1}{5}\right)-\frac{1}{2}× \left(\frac{1}{5}-\frac{1}{7}\right)-\frac{1}{2}× \left(\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}× \left(\frac{1}{9}-\frac{1}{11}\right)$
$=-\frac{1}{2}× \left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)$
$=-\frac{1}{2}× \left(1-\frac{1}{11}\right)$
$=-\frac{1}{2}× \frac{10}{11}=-\frac{5}{11}$.

答案：
(1)$\frac{1}{6}$ $\frac{1}{12}$ $\frac{1}{90}$ $\frac{1}{6}$ $\frac{1}{12}$ $\frac{1}{90}$
(2)$\frac{9}{10}$
(3)$\frac{1}{3× 7}+\frac{1}{7× 11}+\frac{1}{11× 15}+\cdots +\frac{1}{55× 59}$
$=\frac{1}{4}× \left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{4}× \left(\frac{1}{7}-\frac{1}{11}\right)+\frac{1}{4}× \left(\frac{1}{11}-\frac{1}{15}\right)+\cdots +$
$\frac{1}{4}× \left(\frac{1}{55}-\frac{1}{59}\right)$
$=\frac{1}{4}× \left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\cdots +\frac{1}{55}-\frac{1}{59}\right)$
$=\frac{1}{4}× \left(\frac{1}{3}-\frac{1}{59}\right)=\frac{14}{177}$.
(4)$-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}$
$=-\frac{1}{1× 3}-\frac{1}{3× 5}-\frac{1}{5× 7}-\frac{1}{7× 9}-\frac{1}{9× 11}$
$=-\frac{1}{2}× \left(1-\frac{1}{3}\right)-\frac{1}{2}× \left(\frac{1}{3}-\frac{1}{5}\right)-\frac{1}{2}× \left(\frac{1}{5}-\frac{1}{7}\right)-\frac{1}{2}×$
$\left(\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}× \left(\frac{1}{9}-\frac{1}{11}\right)$
$=-\frac{1}{2}× \left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)$
$=-\frac{1}{2}× \left(1-\frac{1}{11}\right)$
$=-\frac{1}{2}× \frac{10}{11}=-\frac{5}{11}$.