a= 1/1.7+4/7.13+4/13,19+4/19.25 +...+4/43.49
K
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CN
1
16 tháng 2 2017
ta có:
\(S=\frac{4}{6}\left(\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{43.49}\right)\)\(=\frac{4}{6}\left(\frac{7-1}{1.7}+\frac{13-7}{7.13}+...+\frac{49-43}{43.49}\right)=\frac{4}{6}\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{43}-\frac{1}{49}\right)\)
\(\frac{4}{6}\left(1-\frac{1}{49}\right)=\frac{4.48}{6.49}=\frac{32}{49}\)
LT
1
25 tháng 6 2015
1/7+1/91+1/247+1/475+1/775+1/1147=? (1)
ta có: (1) <=>: 1/(1.7)+1/(7.13)+1/(13.19)+1/(19.25)+1/(25.31)+1/(31.37)
=1/6.(1-1/7+1/7-1/13+1/13-1/19+1/19-1/25+1/25-1/31+1/31-1/37)
=1/6.(1-1/37)=6/37
G
3 tháng 9 2020
\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(=17.\frac{1}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(=\frac{17}{3}.\frac{6}{49}\)
\(=\frac{34}{49}\)
\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)
\(=13.\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)
\(=13.\frac{1}{3}.\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)\)
\(=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)
\(=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(=\frac{13}{3}.\frac{6}{49}\)
\(=\frac{26}{49}\)
Câu C sai đề rồi , phải là như thế này :
\(C=\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.23}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}.\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{253.258}\right)\)
\(=\frac{4}{5}.\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}.\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}.\frac{125}{1032}\)
\(=\frac{25}{258}\)
29 tháng 3 2017
a, 2016.(-4) - (-1008).8= (-1008).8 - (-1008).8=0
b, Sai đề à bạn
22 tháng 7 2017
E = \(\frac{36}{1\cdot7}+\frac{36}{7\cdot13}+...+\frac{36}{94\cdot100}=\frac{36}{6}\left[\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+...+\frac{1}{94\cdot100}\right]\)
\(=6\left[1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{94}-\frac{1}{100}\right]=6\left[1-\frac{1}{100}\right]\)
\(=6\cdot\frac{99}{100}=\frac{297}{50}\)
F = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)
\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3a+2}-\frac{1}{3a+5}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3a+5}\right]=\frac{1}{6}-\frac{1}{9a+15}\)
G = \(\frac{1}{2\cdot3}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{4}{8\cdot12}+\frac{5}{12\cdot17}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{12}-\frac{1}{17}\)
\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)
NN
1 tháng 4 2019
\(\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+...+\frac{1}{61.67}\)
=6.\(\left(\frac{1}{1.7}+\frac{1}{7.13}+...+\frac{1}{61.67}\right)\):6
=\((\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{61.67}):6\)
=\(\left(1-\frac{1}{7}+\frac{1}{7}+\frac{1}{13}+...+\frac{1}{61}+\frac{1}{67}\right):6\)
=\(\left(1-\frac{1}{67}\right):6\)
=\(\frac{66}{67}:6=\frac{66}{67}.\frac{1}{6}=\frac{11}{67}\)
12 tháng 5 2022
Bài 1:
\(S=4\left(\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+...+\dfrac{1}{43\cdot49}\right)\)
\(=\dfrac{4}{6}\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+...+\dfrac{6}{43\cdot49}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{43}-\dfrac{1}{49}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{48}{49}=\dfrac{96}{147}=\dfrac{32}{49}\)
Bài 3:
Theo đề, ta có:
\(\dfrac{a}{b}=\dfrac{a+10}{b+10}\)
=>ab+10a=ab+10b
=>10a=10b
=>a/b=1
\(=\dfrac{1}{7}+\dfrac{4}{6}\left(\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+...+\dfrac{6}{43\cdot49}\right)\)
\(=\dfrac{1}{7}+\dfrac{2}{3}\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{43}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{7}+\dfrac{2}{3}\cdot\dfrac{6}{49}=\dfrac{1}{7}+\dfrac{4}{49}=\dfrac{11}{49}\)