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
TA
1
23 tháng 2 2023
SỬa đề: \(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{23\cdot27}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\)
=1/3-1/27
=8/27
Y
29 tháng 6 2017
Ta có :
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{23.27}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+....+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}==\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)
DA
16 tháng 8 2017
a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)
\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)
\(\frac{1}{4}xA=\frac{127}{384}\)
\(A=\frac{127}{384}:\frac{1}{4}\)
\(A=\frac{127}{96}\)
KT
15 tháng 10 2018
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
DD
Đoàn Đức Hà
Giáo viên
8 tháng 7 2021
\(A=\frac{2}{3\times7}+\frac{2}{7\times11}+\frac{2}{11\times15}+...+\frac{2}{99\times103}\)
\(2\times A=\frac{4}{3\times7}+\frac{4}{7\times11}+\frac{4}{11\times15}+...+\frac{4}{99\times103}\)
\(=\frac{7-3}{3\times7}+\frac{11-7}{7\times11}+\frac{15-11}{11\times15}+...+\frac{103-99}{99\times103}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{99}-\frac{1}{103}\)
\(=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow A=\frac{50}{309}\)
TH
0
11 tháng 8 2015
\(A=\frac{1}{3\times5}+\frac{3}{5\times11}+\frac{2}{11\times15}+\frac{3}{15\times21}\)
\(A=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{6}{5\times11}+\frac{4}{11\times15}+\frac{6}{15\times21}\right)\)
\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{21}\right)\)
\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(A=\frac{1}{2}\times\frac{2}{7}\)
\(A=\frac{1}{7}\)
HT
11 tháng 8 2015
A = \(\frac{1}{3.5}+\frac{3}{5.11}+\frac{2}{11.15}+\frac{3}{15.21}\)
A = \(\frac{1}{2}.\left(\frac{2}{3.5}+\frac{6}{5.11}+\frac{4}{11.15}+\frac{6}{15.21}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{21}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)
A = \(\frac{1}{2}.\frac{2}{7}\)
A = \(\frac{1}{7}\)
22 tháng 5 2018
a)\(\left(\frac{25}{49}+\frac{17}{39}+\frac{22}{39}\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\left(\frac{17}{39}+\frac{22}{39}\right)\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\frac{39}{39}\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+1\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\frac{50}{49}\times\frac{41}{25}\)
\(=\frac{2050}{1225}\)
B=2/11x15+2/15x19+2/19x23+...+2/51x55
=>2B=4/11x15+4/15x19+4/19x23+...+4/51x55
=1/11-1/15+1/15-1/19+...+1/51-1/55
=1/11-1/55
=5/55-1/55
=4/55
=>B=4/55 : 2
=4/55.1/2
=2/55
\(B=\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\right)\)
\(B=\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{1}{2}.\frac{4}{55}=\frac{2}{55}\)