5/3.7 + 5/7.11 + .........+ 5/85.89
2/15 + 2/35 +2/63 +2/99 + 2/143
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\(K=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{85}-\dfrac{1}{89}\right)\)
\(=\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{89}\right)=\dfrac{5}{4}\cdot\dfrac{86}{267}=\dfrac{215}{534}\)
\(A=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{81\cdot85}+\frac{5}{85\cdot89}\\ A=\frac{5}{4}\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{81\cdot85}+\frac{4}{85\cdot89}\right)\\ A=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{81}-\frac{1}{85}+\frac{1}{85}-\frac{1}{89}\right)\\ A=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{89}\right)\\ A=\frac{5}{4}\left(\frac{89}{267}-\frac{3}{267}\right)\\ A=\frac{5}{4}\cdot\frac{86}{267}=\frac{215}{534}\)
\(5-\dfrac{2}{3}-\dfrac{14}{15}+\dfrac{1}{35}-\dfrac{62}{63}-\dfrac{98}{99}-\dfrac{142}{143}\)
\(=5-\left(1-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{15}\right)+\dfrac{1}{35}-\left(1-\dfrac{1}{63}\right)-\left(1-\dfrac{1}{99}\right)-\left(1-\dfrac{1}{143}\right)\)
\(=5-1+\dfrac{1}{1\cdot3}-1+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}-1+\dfrac{1}{7\cdot9}-1+\dfrac{1}{9\cdot11}-1+\dfrac{1}{11\cdot13}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=1-\dfrac{1}{13}=\dfrac{12}{13}\)
a.\(A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)
\(=\frac{1}{3}-\frac{1}{111}=\frac{37}{111}-\frac{1}{111}=\frac{36}{111}=\frac{12}{37}\)
Vậy A=\(\frac{12}{37}\)
b.\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}=\frac{2}{7}\)
Vậy \(B=\frac{2}{7}\)
c.\(C=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
\(\Rightarrow C.\frac{1}{2}=\left(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\right).\frac{1}{2}\)
\(=\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{4}-\frac{1}{16}=\frac{4}{16}-\frac{1}{16}=\frac{3}{16}\)
Vậy \(C=\frac{3}{16}\)
A = \(\frac{4}{3.7}+\frac{4}{7.9}+...+\frac{4}{107.111}\)
A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{107}-\frac{1}{111}\)
A = \(\frac{1}{3}-\frac{1}{111}\)=\(\frac{12}{37}\)
2 câu sau tương tự. Mik ngại làm lắm -_-
Bài tập 1:
S=2/15+2/35+2/63+2/99+2/143
\(\Rightarrow\)S=2/3x5 +2/5x 7 +2/7x9 +2/9x11 +2/11x13
\(\Rightarrow\)S=1/3 -1/5 +1/5 - 1/7 +1/7 -1/9 +1/9 -1/11 +1/11 -1/13
\(\Rightarrow\)S=1/3 -1/13
\(\Rightarrow\)S=13/39 -3/39
\(\Rightarrow\)S=10/39
S=3/1.4 +3/4.7+3/7.11 ..........sai đề rồi
Bài 2
A=5/11.16+5/16.21+5/21.26+...+5/61.66
\(\Rightarrow\)A=1/11+1/16+1/16-1/21+1/21-1/26+....+1/61-1/66
\(\Rightarrow\)A=1/11-1/66
\(\Rightarrow\)A=6/66-1/66
\(\Rightarrow\)A=5/66
1: \(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=1/2*10/39
=5/39
2: \(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\right)=\dfrac{5}{2}\cdot\dfrac{10}{11}=\dfrac{50}{22}=\dfrac{25}{11}\)
Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)
\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)
\(A=\frac{5}{39}\)
Câu còn lại cx dựa như vậy nhé bn !
Chúc bn hc tốt <3
=2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13
=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
= 1/3 - 1/13
= 13/39 -3/39
=10/39
\(\dfrac{5}{3.7}+\dfrac{5}{7.11}+...+\dfrac{5}{85.89}\)
= \(\dfrac{5}{4}\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{85.89}\right)\)
= \(\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{85}-\dfrac{1}{89}\right)\)
= \(\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{89}\right)\)
= \(\dfrac{215}{534}\)
----------------------------
\(\dfrac{2}{15}+\dfrac{1}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)
= \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
= \(\dfrac{1}{3}-\dfrac{1}{11}\)
= \(\dfrac{8}{33}\)