\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\left(X-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
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\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
\(\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+\left|x+\dfrac{1}{5.7}\right|+.....+\left|x+\dfrac{1}{197.199}\right|=100x\)
23 tháng 11 2021 lúc 20:45
Tìm x biết :\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{1}{2}\left[\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right]=\dfrac{49}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\Leftrightarrow x=49\)
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Tìm sốtựnhiên x biết: \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{x.\left(x+2\right)}=\dfrac{20}{41}\)
Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{20}{41}\)
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{40}{41}\)
\(\Leftrightarrow1-\dfrac{2}{x+2}=\dfrac{40}{41}\)
\(\Leftrightarrow\dfrac{2}{x+2}=\dfrac{1}{41}\)
Suy ra: x+2=82
hay x=80
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tìm x:
a,\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right).\left(x-1\right)+\dfrac{1}{10}.x=x-\dfrac{9}{10}\)
b,\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right).\left(x-2\right)+x=\dfrac{149}{99}.x-\dfrac{98}{99}\)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+......+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\Leftrightarrow198x=196x+98\\ \Leftrightarrow2x=98\Leftrightarrow x=49\)
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Nguyễn Hoàng Minh cho hỏi 2x + 1 - 1 đâu ra v ạ??
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Cho \(S_n=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\). Kết quả giới hạn của limSn bằng:
A. \(\dfrac{1}{3}\)
B. \(\dfrac{1}{5}\)
C. \(\dfrac{1}{2}\)
D. 0
\(S_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2n}{2n+1}\right)=\dfrac{2n}{2\left(2n+1\right)}\)
-> chọn D
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Xem thêm câu trả lời
Adfrac{1}{10}+dfrac{1}{11}+dfrac{1}{12}+...+dfrac{1}{99}+dfrac{1}{100}1
dfrac{2n+5}{n+2}
Sdfrac{1}{20}+dfrac{1}{21}+dfrac{1}{22}+L+dfrac{1}{29}
Adfrac{7}{4}.left(dfrac{3333}{1212}+dfrac{3333}{2020}+dfrac{3333}{3030}+dfrac{3333}{4242}right)
Adfrac{1}{4.5}+dfrac{1}{5.6}+...+dfrac{1}{48.49}+dfrac{1}{49.50}
Adfrac{17}{1.3}+dfrac{17}{3.5}+dfrac{17}{5.7}+...+dfrac{17}{49.51}
dfrac{2n+5}{3n+1}
left(left|xright|+dfrac{2}{5}right):dfrac{2}{5}1
dfrac{1}{2}+dfrac{1}{4}+dfrac{1}{8}+dfrac{1}{16}+dfra...
Đọc tiếp
\(A=\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(\dfrac{2n+5}{n+2}\)
\(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+L+\dfrac{1}{29}\)
\(A=\dfrac{7}{4}.\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{48.49}+\dfrac{1}{49.50}\)
\(A=\dfrac{17}{1.3}+\dfrac{17}{3.5}+\dfrac{17}{5.7}+...+\dfrac{17}{49.51}\)
\(\dfrac{2n+5}{3n+1}\)
\(\left(\left|x\right|+\dfrac{2}{5}\right):\dfrac{2}{5}=1\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}< 1\)
\(A=\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)
\(5.\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{10}\right)\le\dfrac{x}{20}\le-3\left(\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{5}\right)\left(x\in Z\right)\)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}< 1\)
trả hiểu yêu cầu đề bài là j cả
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Tim x, bt:\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{8}{17}\)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{x+2}\right)=\dfrac{8}{17}\)
\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{8}{17}:\dfrac{1}{2}=\dfrac{16}{17}\)
\(\Rightarrow\dfrac{1}{x+2}=1-\dfrac{16}{17}=\dfrac{1}{17}\)
\(\Rightarrow x+2=17\rightarrow x=15\)
Vậy x = 15
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