NH
5
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TM
13 tháng 4 2015
a) thấy dấu cộng ở trước số 6 thành dấu trừ
b) = 2/ 2 + 2/ 6 + 2/ 12 + 2/ 20 + 2/ 30 + 2/ 42 + 2/ 56 + 2/ 72 + 2/ 90
= 2x ( 1/ 1x2 + 1 / 2x3 + 1/ 3x4 + 1/ 4x5 + 1/ 5x6 + 1/ 6x7 + 1/ 7x8 + 1/ 8x9 + 1/ 9x10 )
= 2x ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +1/5 - 1/6 +.. + 1/8- 1/9 + 1/9 - 1/10 )
=2 x( 1 - 1/10 )
=2 x 9/10 = 18/10 = 9 / 5
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
PC
1
2 tháng 7 2023
M=2/6+2/12+...+2/90
=2(1/6+1/12+...+1/90)
=2(1/2-1/3+1/3-1/4+...+1/9-1/10)
=2*4/10=8/10=4/5
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
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\)
\(A=2\times\dfrac{1}{2}\times\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\right)\)
\(A=2\times\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\right)\)
\(A=2\times\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}\right)\)
\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)
\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{11}\right)\)
\(A=2\times\dfrac{9}{22}\)
\(A=\dfrac{9}{11}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 5 2023
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\)
A = 2\(\times\) ( \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\)+ \(\dfrac{1}{72}\))
A =2\(\times\)( \(\dfrac{1}{1\times2}\)+\(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\))
A = 2 \(\times\) ( \(\dfrac{1}{1}\)- \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\))
A = 2\(\times\)( 1 - \(\dfrac{1}{9}\))
A = 2 \(\times\) \(\dfrac{8}{9}\)
A = \(\dfrac{16}{9}\)
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
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)
\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)
\(=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{10\times11}+\frac{1}{11\times12}\right)\)
\(=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)
\(=2\times\left(1-\frac{1}{12}\right)\)
\(=2\times\frac{11}{12}\)
\(=\frac{11}{6}\)
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{45}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{90}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{90}\right)\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{10}\right)=1-\frac{1}{5}=\frac{1}{4}\)
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{36}+\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{72}+\frac{1}{90}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2}A=\frac{2}{5}\)
\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}\)
\(\Rightarrow A=\frac{2}{5}.2\)
\(\Rightarrow A=\frac{4}{5}\)