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Đặt tổng trân là A
Ta có A=1/2x3+1/3x4+...+1/10x11
A=1/2-1/3+1/3-1/4+...+1/10-1/11
Sau khi rút gọn ta được A=1/2-1/11
A=9/22
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = 1/2 + 5/6 + ... + 89/90
9 - A = 1- 1/2 + 1- 5/6 + ... + 1- 89/90
9 -A = 1/2 + 1/6 + ... + 1/90
9 -A = 1/1.2 + 1/2.3 + ... + 1/9.10
9-A = 1/1 - 1/2 + 1/2 - 1/3 +.... + 1/9 - 1/10
9 - A = 1/1 - 1/10
9- A = 9/10
=> A = 9 - 9/10
=> A = 81/10
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\(\text{1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90}\)
\(\text{= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.1}\)
\(\text{= 1/1 - 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10}\)
\(=1/1-1/10-10/10-1/10-9/10\)
Vậy \(\text{ 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 = 9/10}\)
Sửa đề:
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}\)
\(=\dfrac{3}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-...-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}+\frac{1}{90}\right)\)
\(=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
\(=\frac{9}{10}-\frac{9}{10}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A= \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
= \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{9x10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
=\(1-\frac{1}{10}=\frac{9}{10}\)
\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{9\times10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S1=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(S1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(S1=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(S1=1-\frac{1}{10}\)
\(S1=\frac{9}{10}\)
CHÚC BN HC GIỎI !!!!!!!!!! TỨ DIỆP THẢO
S=\(\frac{1}{1.2}+\frac{1}{2.3}+...............+\frac{1}{9.10}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...............+\frac{1}{9}-\frac{1}{10}\)
=\(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nếu biết cách đăng hình vào câu trả lời thì mk sẽ đăng cách giải và đáp số, chứ mk ngại viết lắm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(N=\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{9.10}\)
\(N=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{9}-\frac{1}{10}\)
\(N=\frac{1}{2}-\frac{1}{10}\)
\(N=\frac{2}{5}\)
\(N=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(N=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(N=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(N=\frac{1}{2}-\frac{1}{10}\)
\(N=\frac{2}{5}\)