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AH
Akai Haruma
Giáo viên
9 tháng 11 2021
Lời giải:
$\frac{A}{2}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}$
$=\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}+\frac{9-8}{8\times 9}+\frac{10-9}{9\times 10}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}$
$=1-\frac{1}{9}=\frac{8}{9}$
$\Rightarrow A=2\times \frac{8}{9}=\frac{16}{9}$
BN
1
H9
HT.Phong (9A5)
CTVHS
16 tháng 10 2023
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\)
\(A=2\times\dfrac{1}{2}\times\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\right)\)
\(A=2\times\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\right)\)
\(A=2\times\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}\right)\)
\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)
\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{11}\right)\)
\(A=2\times\dfrac{9}{22}\)
\(A=\dfrac{9}{11}\)
CT
1
LC
3
8 tháng 8 2020
Bài làm:
Ta có: \(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{66}\)
\(=\frac{1}{1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{11.6}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.1.3}+\frac{1}{2.3.2}+...+\frac{1}{2.11.6}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{12}\right)\)
\(=\frac{1}{2}.\frac{11}{12}\)
\(=\frac{11}{24}\)
NH
8 tháng 8 2020
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)
\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)
\(=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{10\times11}+\frac{1}{11\times12}\right)\)
\(=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)
\(=2\times\left(1-\frac{1}{12}\right)\)
\(=2\times\frac{11}{12}\)
\(=\frac{11}{6}\)
L
3
HT
10 tháng 9 2016
\(a=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(a=\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+\frac{1}{3.5}+...+\frac{1}{5.9}\)
\(a=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(a=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(a=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
=> \(a=2.\frac{2}{5}\)
=> \(a=\frac{4}{5}\)
10 tháng 9 2016
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\right)\cdot\frac{1}{2}\)
\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)
LN
1
BV
23 tháng 12 2016
1/3+1/6+....(bạn viết nốt cho mình cái đề bài mình đang ngại)
suy ra:
1/1*3+1/3*2+1/2*5+....+1/9*5(cứ viết thế này thôi thầy cô đọc sẽ hiểu ngay)
suy ra:1-1/3+1/3-1/2+1/2-1/5+......+1/9-1/5
suy ra:1-1/5=4/5(cậu rút gọn 1/3 cho 1/3,1/2 cho 1/2 cứ như thế sẽ thừa ra 1/5 rồi lấy số 1 thừa ban đầu trừ đi 1/5 là ok)
PC
1
TG
11 tháng 4 2022
1/2 N=1/2x3 + 1/3x4 +...+1/9x10
1/2 N=1/2-1/3+1/3-1/4+...+1/9-1/10
1/2 N=1/2-1/10=2/5
N=2/5:1/2=4/5
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(=2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\right)\)
\(=2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{4}{5}\)
bạn sai rồi
chỉ là 1/6 +1/12 +...+1/90 thôi nhé