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![](https://rs.olm.vn/images/avt/0.png?1311)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{2017}\right)\)
\(=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}\)
\(=\frac{3.4....2018}{2.3....2017}=\)\(\frac{2018}{2}=1009\)
=\(\frac{3}{2} × \frac{4}{3} × \frac{5}{4} ....× \frac{2018}{2017}\)
=>2008/2 = 1004
K MK NHA. CHÚC BẠN HỌC GIỎI
![](https://rs.olm.vn/images/avt/0.png?1311)
= (1/2).(2/3).(4/5).(5/6)......(2016/2017).(2017/2018)
=1.2.3.4.5......2016.2017/2.3.4.5.....2017.2018
=1/2018
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2016\cdot2017}{2\cdot3\cdot4\cdot\cdot\cdot\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)
2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)
Vậy ......
hok tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
các bạn ơi mk đánh nhầm dấu cộng là dấu nhân nhé
\(\left(1-\frac{4}{1}\right).\left(1-\frac{4}{9}\right)...\left(1-\frac{4}{2017^2}\right)\)
\(=\left(\frac{1-4}{1}\right).\left(\frac{9-4}{9}\right)...\left(\frac{2017^2-4}{2017^2}\right)\)
\(=\left(-3\right).\left(\frac{1.5}{3.3}\right).\left(\frac{3.7}{5.5}\right)...\left(\frac{2015.2019}{2017.2017}\right)\)
\(=-\left(3\right).\frac{1}{3}.\frac{2019}{2017}=-\frac{2019}{2017}\)