Tính tổng A=\(\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+...+\dfrac{5}{2019.2023}\)
các bạn làm đầy đủ các bước đừng làm tắt giúp mình nhé.Mình cảm ơn ạ!
Tính tổng A= 5/3.7 + 5/7.11 + 5/11.15+...+5/2019.2023
Các bạn cho mình hỏi mình viết thành 5/4 x (4/3.7 + 4/7.11 + 4/11.15+...+ 4/2019.2023 ) được ko ạ?
Viết vậy đúng đó em
A = 5/(3.7) + 5/(7.11) + 5/(11.15) + ... + 5/(2019.2023)
= 5/4 . [4/(3.7) + 4/(7.11) + 4/(11.15) + ... + 4/(2019.2023)]
= 5/4 . (1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/2019 - 1/2023)
= 5/4 . (1/3 - 1/2023)
= 5/4 . 2020/6069
= 2525/6069
tính tổng sau : \(K=\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+...+\dfrac{5}{81.85}+\dfrac{5}{85.89}\)
\(K=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{85}-\dfrac{1}{89}\right)\)
\(=\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{89}\right)=\dfrac{5}{4}\cdot\dfrac{86}{267}=\dfrac{215}{534}\)
CM: \(\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+.....+\dfrac{3}{\left(4n-1\right)\left(4n+3\right)}=\dfrac{5n}{4n+3}\)
\(\dfrac{5}{3\cdot7}+\dfrac{5}{7\cdot11}+\dfrac{5}{11\cdot15}+...+\dfrac{5}{\left(4n-1\right)\left(4n+3\right)}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\dfrac{4n}{12n+9}\\ =\dfrac{5n}{12n+9}\)
Mk thực sự nghĩ đề hình như bị sai hay sao ấy!
So sánh tổng S với 251
S = \(\dfrac{1}{2}-\dfrac{1}{3.7}-\dfrac{1}{7.11}-\dfrac{1}{11.15}-\dfrac{1}{15.19}-\dfrac{1}{19.23}-\dfrac{1}{23.27}\)
Mai mk thi r cho mình xem cách làm bài này nhé. Giúp mình với. HELP ME !!!
a) \(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{2011.2013}\)
b) \(\dfrac{1}{7.11}\)+\(\dfrac{1}{11.15}\)+\(\dfrac{1}{15.19}\)+...\(\dfrac{1}{2019.2023}\)
a) 1/(5.7) + 1/(7.9) + ... + 1/(2011.2013)
= 1/2.(1/5 - 1/7 + 1/7 - 1/9 + ... + 1/2011 - 1/2013)
= 1/2.(1/5 - 1/2013)
= 1/2 . 2008/10065
= 1004/10065
b) 1/(7.11) + 1/(11.15) +1/(15.19) + ... + 1/(2019.2023)
= 1/4.(1/7 - 1/11 + 1/11 - 1/15 + 1/15 - 1/19 + ... + 1/2019 - 1/2023)
= 1/4.(1/7 - 1/2023)
= 1/4 . 288/2023
= 72/2023
tính các tổng sau bằng cách hợp lí
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2021.2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+......+\dfrac{4}{107.111}\)
\(S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+.....+\dfrac{1}{60}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)
Tìm phân số \(\dfrac{a}{b}\):
\(\dfrac{2}{7}\) x \(\dfrac{a}{b}\) + \(\dfrac{a}{b}\) x \(\dfrac{5}{7}\) = \(\dfrac{5}{7}\)
các bạn làm nhanh giúp mik nhé! mik cảm ơn
\(\Rightarrow\dfrac{a}{b}\times\left(-\dfrac{2}{7}+\dfrac{5}{7}\right)=\dfrac{5}{7}\\ \Rightarrow\dfrac{a}{b}\times\dfrac{3}{7}=\dfrac{5}{7}\\ \Rightarrow\dfrac{a}{b}=\dfrac{5}{7}:\dfrac{3}{7}\\ \Rightarrow\dfrac{a}{b}=\dfrac{5}{3}\\ \Rightarrow a=5;b=3\)
\(=>\dfrac{a}{b}\times\left(-\dfrac{2}{7}+\dfrac{5}{7}\right)=\dfrac{5}{7}\)
\(=>\dfrac{a}{b}\times\dfrac{3}{7}=\dfrac{5}{7}=>\dfrac{a}{b}=\dfrac{5}{7}:\dfrac{3}{7}=\dfrac{5}{3}\)
vậy \(\dfrac{a}{b}=\dfrac{5}{3}\)
Tính: \(\dfrac{5}{12}+\dfrac{5}{20}+\dfrac{5}{30}+\dfrac{5}{42}+\dfrac{5}{56}\)
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\(=5\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\right)\)
\(=5\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)\)
\(=5\cdot\dfrac{5}{24}=\dfrac{25}{24}\)
`5/12 + 5/20 + 5/30 + 5/42 + 5/56`
`= 5/(3.4) + 5/(4.5) + 5/(5.6) + 5/(6.7) + 5/(7.8)`
`= 5 . (1/(3.4) + 1/(4.5) + ... + 5/(7.8))`
`= 5 . (1/3 - 1/4 + 1/4 - 1/5 + ...+ 1/7 - 1/8)`
`= 5 . (1/3 - 1/8) `
`= 5 . 5/24 = 25/24`
làm đầy đủ theo các bước nhé
Tìm x biết :
a) \(^{\dfrac{4}{9}+x=\dfrac{5}{3}}\)
b)\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
c) \(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
d)\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)
\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)
\(x=-\dfrac{15}{2}\)
d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)
\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)
A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)
\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)
\(x\)\(=\dfrac{11}{9}\)
B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)
\(x=\)\(\dfrac{-2}{3}\)
a)
\(\frac{4}{9} + x = \frac{5}{3}\)
=> \(x = \frac{5}{3}-\frac{4}{9}\)
=> \(x = \) \(\frac{11}{9}\)
Vậy \(x = \dfrac{11}{9}\)
b)
\(\dfrac{3}{4} .x = \dfrac{-1}{2}\)
=> \(x = \dfrac{-1}{2} : \dfrac{3}{4}\)
=> \(x = \dfrac{-2}{3}\)
Vậy \(x = \dfrac{-2}{3}\)
c)
\( \dfrac{3}{7}+ \dfrac{5}{7}:x = \dfrac{1}{3}\)
=> \(\dfrac{5}{7}:x = \dfrac{1}{3}-\) \( \dfrac{3}{7}\)
=> \(\dfrac{5}{7}:x = \dfrac{-2}{21}\)
=> \(x = \dfrac{5}{7}:\dfrac{-2}{21}\)
=> \(x = \dfrac{-15}{2}\)
Vậy \(x = \dfrac{-15}{2}\)
d)
\(3\dfrac{1}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)
=> \(\dfrac{13}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)
=> \( |2x - \dfrac{5}{12} | =\dfrac{13}{4} : \dfrac{39}{16}\)
=> \(|2x-\dfrac{5}{12} |= \dfrac{4}{3}\)
=> \(\left[\begin{matrix} 2x - \dfrac{5}{12} = \dfrac{4}{3}\\ 2x - \dfrac{5}{12} = \dfrac{4}{3}\end{matrix}\right.\)
=> \(\left[\begin{matrix} 2x = \dfrac{-4}{3}+\dfrac{5}{12}\\ 2x = \dfrac{-4}{3}+\dfrac{5}{12} \end{matrix}\right.\)
=> \(\left[\begin{matrix} 2x = \dfrac{7}{4}\\ 2x = \dfrac{-11}{12} \end{matrix}\right.\)
=> \(\left[\begin{matrix} x = \dfrac{7}{8}\\ x = \dfrac{-11}{24} \end{matrix}\right.\)
Vậy \(x \in \) { \(\dfrac{7}{8} ; \dfrac{-11}{24}\) }