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\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{9\cdot11}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{9\cdot11}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{11-9}{9\cdot11}\right)\)
\(=\frac{1}{2}\left(\frac{3}{1\cdot3}-\frac{1}{1\cdot3}+\frac{5}{3\cdot5}-\frac{3}{3\cdot5}+...+\frac{7}{5\cdot7}-\frac{5}{5\cdot7}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(=\frac{1}{2}\cdot\frac{10}{11}\)
\(=\frac{10}{22}=\frac{5}{11}\)
Ta có :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(=\)\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\)\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\)\(\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(=\)\(\frac{1}{2}.\frac{10}{11}\)
\(=\)\(\frac{5}{11}\)
Bạn làm đúng òi
Chúc bạn học tốt ~
- \(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\)
\(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)
\(4.B=1-\frac{1}{97}\)
\(4.B=\frac{96}{97}\)
\(B=\frac{96}{97}:4\)
\(B=\frac{24}{97}\)
a)
\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}\)
b)\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5}{20}-\dfrac{4}{20}=\dfrac{1}{20}\)
c) Ta có: \(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{n\cdot\left(n+1\right)}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}\)
\(=\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\)
M=2/3.5+2/5.7+...+2/97.99
M=1/3-1/5+1/5-...+1/97-1/99
M=1/3-1/99
M=32/99
\(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{97.99}\)
\(=2\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-...-\frac{1}{97}+\frac{1}{99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=2.\frac{32}{99}\)
\(=\frac{64}{99}\)
Mk bik câu B nè!
2B = 2/3.5 + 2/5.7 + 2/7.9 +.......+2/97.99
2B = 1/3 - 1/5 + 1/5 - 1/7 +.......+ 1/97 - 1/99
2B = 1/3 - 1/99
2B = 32/99
=> B = 16/99
✫¸.•°*”˜˜”*°•✫ Ṱђầภ Ḉђết ✫•°*”˜˜”*°•.¸✫ nhân A với 2 rồi phân tích như vậy được
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+....+\frac{1}{37\cdot39}\)
\(2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+....+\frac{2}{37\cdot39}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{37}-\frac{1}{39}\)
\(2A=\frac{1}{3}-\frac{1}{39}=\frac{12}{39}=\frac{4}{13}\)
\(A=\frac{4}{13}:2=\frac{4}{13}\cdot\frac{1}{2}=\frac{2}{13}\)
Vậy \(A=\frac{2}{13}\)
Bài làm
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(A=\frac{1}{3}-\frac{1}{39}\)
\(A=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)
Vậy \(A=\frac{12}{39}\)
=>2A=1/3-1/5+1/5-1/7+1/7-1/9+...+1/37-1/39
=>2A=1/3-1/39=4/13
=>A=2/13
\(M=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(\Rightarrow M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(\Rightarrow M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(\Rightarrow M=\frac{1}{2}.\frac{32}{99}\)
\(\Rightarrow M=\frac{16}{99}\)
\(M=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5.}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{16}{99}\)