H= 1/6 + 1/12 + 1/20 + 1/30 = 1/42 + 1/72 + 1/90
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HM
8 tháng 9 2016
em lớp 6 nha
B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72
B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
B=1+0-0-0-0-0-0-0-1/9
B=1-1/9
B=8/9
k em nha
NA
Nguyễn An Ninh
VIP
13 tháng 5 2023
\(\text{1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90}\)
\(\text{= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.1}\)
\(\text{= 1/1 - 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10}\)
\(=1/1-1/10-10/10-1/10-9/10\)
Vậy \(\text{ 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 = 9/10}\)
KV
13 tháng 5 2023
Sửa đề:
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}\)
\(=\dfrac{3}{5}\)
PE
1
14 tháng 8 2015
A=1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
A=1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10
A=1/2-1/10
A=2/5
3 tháng 9 2016
\(\text{Ta có: }\)\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{9.8}+\frac{1}{8.7}+\frac{1}{7.6}+\frac{1}{6.5}+\frac{1}{5.4}+\frac{1}{4.3}+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=-\frac{81}{90}=-\frac{9}{10}\)
3 tháng 9 2016
<=>
D = 1/90+1/72+1/56+1/42+1/30+1/20+1/12+1/6+1/2
D = 1/(1x2) + 1/(2x3) + 1/(3x4) + 1/(4x5) + 1/(5x6) + … + 1/(9x10)
D = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + …. + 1/9 – 1/10
D = 1 – 1/10
D = 9/10
11 tháng 9 2020
( 1/90 + 1/72 + ... + 1/2)
= - { 1/( 9.10) + 1/(9.8) + ... + 1/( 2.1)}
= - ( 1/9 - 1/10 + 1/8 - 1/9 + ...+ 1 - 1/2) ( 1/90 = 1/(9.10) = 1/9 - 1/10)
= - ( 1 - 1/10)
= -9/10
NM
2
30 tháng 6 2020
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+.....+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
14 tháng 7 2021
Bn ghi đề sai nên mik sửa nha!mik từng làm rồi ko sai đâu
B=-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6
B=-(1/90+1/56+1/42+1/30+1/20+1/12+1/6)
B=-(1/10.9+1/8.9+1/8.7+1/7.6+1/6.5+1/5.4+1/4.3+1/3.2)
B=-(1/10-1/9+1/9-1/8+1/8-1/7+1/7-1/6+1/6-1/5+1/5-1/4+1/4-1/3+1/3-1/2)
B=-(1/10-1/2)
B=2/5
HẾT
DT
12
28 tháng 4 2015
\(A=\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{10}-\frac{1}{10}\right)\)\(=\left(\frac{1}{2}-\frac{1}{11}\right)+0+...+0=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
NN
3
11 tháng 8 2016
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{11}\)
\(=\frac{9}{22}\)
11 tháng 8 2016
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=\left(\frac{1}{2}-\frac{1}{11}\right)+0+...+0\)
\(A=\frac{11}{22}-\frac{2}{22}\)
\(A=\frac{9}{22}\)
Sửa đề:
\(H=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\\ H=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\\ H=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ H=\dfrac{1}{2}-\dfrac{1}{10}\\ H=\dfrac{2}{5}\)
Vậy \(H=\dfrac{2}{5}\)
\(H=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(H=\dfrac{1}{90}+\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{72}\right)\)
\(H=\dfrac{1}{90}+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\right)\)
\(H=\dfrac{1}{90}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\right)\)
\(H=\dfrac{1}{90}+\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\)
\(H=\dfrac{1}{90}+\left(\dfrac{4}{8}-\dfrac{1}{8}\right)\)
\(H=\dfrac{1}{90}+\dfrac{3}{8}\)
\(H=\dfrac{49}{360}\)