Cho M=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
So sanh M vs 1/2
Những câu hỏi liên quan
bai 1: s=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
bai 2: s=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\text{+}...\text{+}\frac{1}{2017.2019}\)
co ban nao ra chua de minh do ke qua coi dung ko?
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\(S=\frac{1}{1.2}+\frac{1}{3.4}+.........+\frac{1}{199.200}\)
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Bài 2:
\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2017.2019}\)
\(=\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2017.2019}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\cdot\frac{2018}{2019}=\frac{1009}{2019}\)
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Mk cần gấp lắm!Ai nhanh thì mk cho 3 tích!Tính tổng sau: F=\(\frac{8}{1.3}+\frac{8}{3.5}+\frac{8}{5.7}+...+\frac{8}{2017.2019}\) H=\(\frac{1}{8.7}+\frac{1}{14.10}+\frac{1}{20.13}+\frac{1}{26.16}+\frac{1}{32.19}+\frac{1}{38.22}\)
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tìm x, biết
\(\frac{1}{2}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-\frac{1}{9.11}=\frac{4}{5}-x\)
\(\frac{1}{2}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-\frac{1}{9.11}=\frac{4}{5}-x\)
<=> \(2.\frac{1}{2}-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)=\frac{8}{5}-2x\)
<=> \(1-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)=\frac{8}{5}-2x\)
<=> \(1-\left(1-\frac{1}{11}\right)-\frac{8}{5}=-2x\)
<=> \(-\frac{83}{55}=-2x\)
<=> \(x=\frac{83}{110}\)
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tìm x,biết
\(\frac{1}{2}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-\frac{1}{9.11}=\frac{4}{5}-x\)
tính nhanh
M=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{995.997}+\frac{1}{997.999}\)
=> 2M = 2/1*3 + 2/3*5 + ... + 2/995*997 + 2/ 997*999 = 1-1/3 + 1/3 - 1/5 +... + 1/ 995 - 1/997 + 1/997 - 1 / 999 = 1- 1/999 = 998/999
=> m = 998/999 / 2 = 499/999 (bn tính lại xem nha mk ko có máy tính nên sợ sai)
Vậy M = .....
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\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{995.997}+\frac{1}{997.999}\)
\(M=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{995}-\frac{1}{997}+\frac{1}{997}-\frac{1}{999}\)
\(M=1-\frac{1}{999}\)
\(M=\frac{998}{999}\)
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Xem thêm câu trả lời
Tính M=\(\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+...+\frac{99^2}{197.199}\)
Tính
a)S1=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
b)S2=\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
c)S3=\(\frac{1}{10.9}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405}\)
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
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a,tính \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{19.21}\)
b,CMR A\(=\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{(2n-1).(2n+1)}\le\frac{1}{2}\)
tớ làm câu b thôi, câu a nhân 1/2 lên là đc
\(A=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\right)\right]\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2.n-1}-\frac{1}{2n+1}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2n+1}\right)=\frac{1}{2}-\frac{1}{2.\left(2n+1\right)}< \frac{1}{2}\)
p/s: lưu ý không có dấu "=" đâu nhé vì \(\frac{1}{2.\left(2n+1\right)}>0\left(n\text{ thuộc }N\right)\)
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Tính tổng: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
giúp mik vs, ai nhanh tick luôn nè:)))
=49/99 NHA
HT
k cho mình nha
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\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
2A=\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
2A=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)
2A=\(1-\frac{1}{99}\)
A=\(\frac{49}{99}\)
Chúc bạn học tốt
HYC-30/1/2022