a=1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72
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a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)
b) Bạn làm tương tự.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{38}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`A=1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90`
`=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90`
`=9-(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)`
`=9-(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/(9.10))`
`=9-(1-1/2+1/2-1/3+.....+1/9-1/10)`
`=9-(1-1/10)`
`=9-1+1/10=8+1/10=81/10`
A = \(\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+...+\left(1-\dfrac{1}{90}\right)\)
= \(9-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\)
=\(9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
= \(9-\left(1-\dfrac{1}{10}\right)\)
= \(9-\dfrac{9}{10}=\dfrac{81}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)
Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1
\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)
Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72=(1-1/2)+(1-1/6)+(1-1/12)+(1-1/20)+(1-1/30)+(1-1/42)+(1-1/56)+(1-1/72)=(1+1+1+1+1+1+1+1)-(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72)=8-(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)=8-(1-1/2+1/2-1/3+...+1/8-1/9)=8-(1-1/9)=8-8/9=72/9-8/9=64/9
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72
=10-(1/2+1/6+..+1/110)
=10-(1/1x2+1/2x3+...+1/10x11)
=10-(1-1/2+1/2-1/3+...+1/10-1/11)
=10-(1-1/11)
=10-10/11
=100/11
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+..+\left(1-\frac{1}{90}\right)\)
\(A=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
Đặt biểu thức trong ngoặc đơn là B
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(B=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{10-9}{9.10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)
\(A=9-B=9-\frac{9}{10}=8\frac{1}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
9 - A = 1 - 1/2 + 1-5/6 + 1 - 11/12 + ... + 1-89/90
9 - A = 1/2 + 1/6 + 1/12 + .. + 1/90
9 - A = 1/1.2 + 1/2.3 + 1/3.4 + .. + 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
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
\(B=\frac{1}{2}+\frac{1}{6}+................+\frac{1}{90}=\frac{1}{1x2}+.....+\frac{1}{9x10}=1-\frac{1}{2}+...+\frac{1}{9}-\frac{1}{10}=\frac{9}{10}\)
Dễ thấy B+A=1+1+...+1=10
=>A=10-B=\(10-\frac{9}{10}=\frac{91}{10}\)
\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+....+\frac{89}{90}\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+..+\left(1-\frac{1}{90}\right)\)
\(A=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+\left(1-\frac{1}{3.4}\right)+...+\left(1-\frac{1}{9.10}\right)\)
\(A=\left(1+1+1+...+1\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
9 số 1
\(A=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9-\left(1-\frac{1}{10}\right)\)
\(A=9-\frac{9}{10}\)
\(A=\frac{81}{10}\)
Ủng hộ mk nha ^-^
\(a=1-\dfrac{1}{2}+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}+1-\dfrac{1}{30}+1-\dfrac{1}{42}+1-\dfrac{1}{56}+1-\dfrac{1}{72} =\left(1+1+1+1+1+1+1+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=8-\left(\dfrac{6}{12}+\dfrac{2}{12}+\dfrac{1}{12}+\dfrac{3}{60}+\dfrac{2}{60}+\dfrac{12}{504}+\dfrac{9}{504}+\dfrac{7}{504}\right)=8-\left(\dfrac{9}{12}+\dfrac{5}{60}+\dfrac{28}{504}\right)=8-\left(\dfrac{3}{4}+\dfrac{1}{12}+\dfrac{1}{18}\right)=8-\left(\dfrac{27}{36}+\dfrac{3}{36}+\dfrac{2}{36}\right)=8-\dfrac{32}{36}=8-\dfrac{8}{9}=\dfrac{72}{9}-\dfrac{8}{9}=\dfrac{64}{9} \)
Đáp số 64/9
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