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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

\(\Rightarrow A=\frac{2}{4}\left(\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+.....+\frac{1}{51.55}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{1}{2}.\frac{4}{55}=\frac{2}{55}\)

\(VậyA=\frac{2}{55}\)

DD
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. 

10 tháng 12 2021

\(\overline{abcabc}=1001.\overline{abc}=7.11.13.\overline{abc}\)

7, 11, 13 là các số nguyên tố

7 tháng 9 2018

=\(1\left(\frac{1}{14.15}+\frac{1}{15.19}+......+\frac{1}{51.55}\right)\)

=\(1\left(\frac{1}{14}-\frac{1}{15}\right)+\left(\frac{1}{15}-\frac{1}{19}\right).....+\left(\frac{1}{51}-\frac{1}{55}\right)\)

=\(1\left(\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}....+\frac{1}{51}-\frac{1}{55}\right)\)

=\(1\left(\frac{1}{14}-\frac{1}{55}\right)\)

=\(1.\frac{41}{770}\)

=\(\frac{41}{770}\)

20 tháng 3 2020

Bạn tham khảo bài làm của mình nhé !!

\(A=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.23}...+\frac{2}{51.55}\)

\(\Leftrightarrow2A=\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\)

\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\)

\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{55}\)

\(\Leftrightarrow2A=\frac{4}{55}\)

\(\Leftrightarrow A=\frac{4}{110}\)

20 tháng 3 2020

thank nhưng tui làm được rồi

\(A=\dfrac{-5}{3}\cdot\dfrac{11}{2}\cdot\dfrac{4}{3}=\dfrac{-20\cdot11}{2\cdot9}=\dfrac{-110}{9}\)

\(B=\dfrac{2}{4}\left(\dfrac{4}{11\cdot15}+\dfrac{4}{15\cdot19}+...+\dfrac{4}{51\cdot55}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)

=1/2*4/55

=2/55

AH
Akai Haruma
Giáo viên
12 tháng 1 2019

Lời giải:

\(A=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.23}+...+\frac{2}{51.55}\)

\(\Rightarrow 2A=\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\)

\(=\frac{15-11}{11.15}+\frac{19-15}{15.19}+\frac{23-19}{19.23}+....+\frac{55-51}{51.55}\)

\(=\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+...+\frac{1}{51}-\frac{1}{55}\)

\(=\frac{1}{11}-\frac{1}{55}=\frac{4}{55}\)

\(\Rightarrow A=\frac{2}{55}\)

15 tháng 7 2017

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}+0+0+0+0\)

\(=\frac{8}{27}\)

15 tháng 7 2017

Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

Câu 1: 

a: \(A=\dfrac{1}{2}\left(\dfrac{4}{11\cdot15}+\dfrac{4}{15\cdot19}+...+\dfrac{4}{51\cdot55}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{4}{55}=\dfrac{2}{55}\)

\(B=\dfrac{-5}{3}\cdot\dfrac{11}{2}\cdot\dfrac{4}{3}=\dfrac{-220}{18}=\dfrac{-110}{9}\)

\(A\cdot B=\dfrac{2}{55}\cdot\dfrac{-110}{9}=\dfrac{-4}{9}\)

Câu 2: 

a: |3-x|=x-5

=>|x-3|=x-5

\(\Leftrightarrow\left\{{}\begin{matrix}x>=5\\\left(x-5-x+3\right)\left(x-5+x-3\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)