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2/11.15+2/15.19+......+2/51.55
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17 tháng 6 2022
\(=\dfrac{2}{4}\left(\dfrac{4}{11\cdot15}+\dfrac{4}{15\cdot19}+...+\dfrac{4}{51\cdot55}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{4}{55}=\dfrac{2}{55}\)
BC
2
MH
6 tháng 2 2016
\(A=\frac{1}{2}.\left(\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+...+\frac{1}{51}-\frac{1}{55}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{1}{2}.\frac{4}{55}=\frac{2}{55}\)
6 tháng 2 2016
\(\Rightarrow A=\frac{2}{4}\left(\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+.....+\frac{1}{51.55}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{1}{2}.\frac{4}{55}=\frac{2}{55}\)
\(VậyA=\frac{2}{55}\)
LT
1
NH
20 tháng 3 2020
Bạn tham khảo bài làm của mình nhé !!
\(A=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.23}...+\frac{2}{51.55}\)
\(\Leftrightarrow2A=\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\)
\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\)
\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{55}\)
\(\Leftrightarrow2A=\frac{4}{55}\)
\(\Leftrightarrow A=\frac{4}{110}\)
23 tháng 3 2023
\(A=\dfrac{-5}{3}\cdot\dfrac{11}{2}\cdot\dfrac{4}{3}=\dfrac{-20\cdot11}{2\cdot9}=\dfrac{-110}{9}\)
\(B=\dfrac{2}{4}\left(\dfrac{4}{11\cdot15}+\dfrac{4}{15\cdot19}+...+\dfrac{4}{51\cdot55}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)
=1/2*4/55
=2/55
AH
Akai Haruma
Giáo viên
12 tháng 1 2019
Lời giải:
\(A=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.23}+...+\frac{2}{51.55}\)
\(\Rightarrow 2A=\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\)
\(=\frac{15-11}{11.15}+\frac{19-15}{15.19}+\frac{23-19}{19.23}+....+\frac{55-51}{51.55}\)
\(=\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+...+\frac{1}{51}-\frac{1}{55}\)
\(=\frac{1}{11}-\frac{1}{55}=\frac{4}{55}\)
\(\Rightarrow A=\frac{2}{55}\)
DD
Đoàn Đức Hà
Giáo viên
10 tháng 12 2021
a) \(A=\frac{2}{11.15}+\frac{2}{15.19}+...+\frac{2}{51.55}\)
\(=\frac{1}{2}\left(\frac{4}{11.15}+\frac{4}{15.19}+...+\frac{4}{51.55}\right)\)
\(=\frac{1}{2}\left(\frac{15-11}{11.15}+\frac{19-15}{15.19}+...+\frac{55-51}{51.55}\right)\)
\(=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\right)\)
\(=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{2}{55}\)
b) \(\overline{abcabc}=\overline{abc}.1001=\overline{abc}.7.11.13\)suy ra đpcm.
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
10 tháng 12 2021
\(\overline{abcabc}=1001.\overline{abc}=7.11.13.\overline{abc}\)
7, 11, 13 là các số nguyên tố
NN
1
7 tháng 9 2018
=\(1\left(\frac{1}{14.15}+\frac{1}{15.19}+......+\frac{1}{51.55}\right)\)
=\(1\left(\frac{1}{14}-\frac{1}{15}\right)+\left(\frac{1}{15}-\frac{1}{19}\right).....+\left(\frac{1}{51}-\frac{1}{55}\right)\)
=\(1\left(\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}....+\frac{1}{51}-\frac{1}{55}\right)\)
=\(1\left(\frac{1}{14}-\frac{1}{55}\right)\)
=\(1.\frac{41}{770}\)
=\(\frac{41}{770}\)
Đặt :
\(A=\dfrac{2}{11.15}+\dfrac{2}{15.19}+.......+\dfrac{2}{51.55}\)
\(\Leftrightarrow2A=\dfrac{4}{11.15}+\dfrac{4}{15.19}+.........+\dfrac{4}{51.55}\)
\(\Leftrightarrow2A=\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+......+\dfrac{1}{51}-\dfrac{1}{55}\)
\(\Leftrightarrow2A=\dfrac{1}{11}-\dfrac{1}{55}\)
\(\Leftrightarrow2A=\dfrac{4}{55}\)
\(\Leftrightarrow A=\dfrac{4}{110}\)