4/3.7+4/7.11+4/11.15+...+4/85.89
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\(K=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{85}-\dfrac{1}{89}\right)\)
\(=\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{89}\right)=\dfrac{5}{4}\cdot\dfrac{86}{267}=\dfrac{215}{534}\)
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
\(A=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{81\cdot85}+\frac{5}{85\cdot89}\\ A=\frac{5}{4}\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{81\cdot85}+\frac{4}{85\cdot89}\right)\\ A=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{81}-\frac{1}{85}+\frac{1}{85}-\frac{1}{89}\right)\\ A=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{89}\right)\\ A=\frac{5}{4}\left(\frac{89}{267}-\frac{3}{267}\right)\\ A=\frac{5}{4}\cdot\frac{86}{267}=\frac{215}{534}\)
a=4/3.7 +4/7.11+4/11.15 +.....+4/107/111
=1/3-1/7+1/7-1/11+1/11-1/15+......+1/107-1/111
=1/3-1/111
=12/37
\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{59.63}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{59}-\dfrac{1}{63}\)
\(=\dfrac{1}{3}-\dfrac{1}{63}\)
\(=\dfrac{20}{63}\)
A = \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)
A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)
A = \(\frac{1}{3}-\frac{1}{99}\)
A = \(\frac{32}{99}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)
\(\Rightarrow=\frac{1}{3}-\frac{1}{111}\)
\(=\frac{12}{37}\)
k nha
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)
\(=\frac{1}{3}-\frac{1}{111}\)
\(=\frac{108}{333}=\frac{12}{37}\)
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{\left(3x-1\right).\left(3x+3\right)}=\frac{3}{10}\)
=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)
=> \(\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
=> \(\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}\)
=> \(\frac{1}{3x+3}=\frac{1}{30}\)
=> 3x + 3 = 30
=> 3.(x + 1) = 30
=> x + 1 = 10
=> x = 9
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}+0+0+0+0\)
\(=\frac{8}{27}\)
Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}\)
\(=\frac{8}{27}\)
ta có :
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+..+\frac{4}{85.89}\)
\(=\frac{7-3}{3.7}+\frac{11-7}{7.11}+...+\frac{89-85}{85.89}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+..+\frac{1}{85}-\frac{1}{89}=\frac{1}{3}-\frac{1}{89}\)
\(=\frac{86}{267}\)
4/3.7+4/7.11+4/11.15+...+4/85.89
=1/3-1/7+1/7-1/11+1/11-1/15+...+1/85-1/89
=1/3-1/89
=86/267