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\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-....-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)
Nên 3x + 3 = 30
3x = 30 - 3 = 27
x = 27 : 3 = 9
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{3\left(x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)(Vì 3x + 3 lớn hơn 3x - 1 là 4 đơn vị)
\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{x+1-1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{x}{3x+3}=\frac{3}{10}\)
\(\Rightarrow10x=3.\left(3x+3\right)\)
\(\Rightarrow10x=9x+9\)
\(\Rightarrow x=9\)
Vậy...
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}=\frac{10}{30}-\frac{9}{30}=\frac{1}{30}\)
\(\Rightarrow\left(3x+3\right).1=1.30\Rightarrow3x+3=30\Rightarrow3x=27\Rightarrow x=9\)
Ta có:
\(C=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{40.41}+\frac{2}{41.42}\)
\(\Rightarrow C=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}+\frac{1}{41.42}\right)\)
\(\Rightarrow C=2\left(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{41-40}{40.41}+\frac{42-41}{41.42}\right)\)
\(\Rightarrow C=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{40}-\frac{1}{41}+\frac{1}{41}-\frac{1}{42}\right)\)
\(\Rightarrow C=2.\left(\frac{1}{3}-\frac{1}{42}\right)=\frac{13}{21}\)
\(D=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(\Rightarrow D=\frac{7-3}{3.7}+\frac{11-7}{7.11}+\frac{15-11}{11.15}+...+\frac{111-107}{107.111}\)
\(\Rightarrow D=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}=\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)\(E=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(\Rightarrow E=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(\Rightarrow E=\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}+\frac{11-10}{10.11}\)
\(\Rightarrow E=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)
a) \(A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot13}+...+\frac{3}{647\cdot650}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{647}-\frac{1}{650}=\frac{1}{5}-\frac{1}{650}=\frac{129}{650}\)
b) \(B=\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{196\cdot200}=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{196\cdot200}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{196}-\frac{1}{200}\right)=3\left(\frac{1}{3}-\frac{1}{200}\right)=3\cdot\frac{197}{600}=\frac{197}{200}\)
sửa 199 -> 200
P/S : Lần sau đừng có đăng từng câu từng câu hỏi trên đây nhá
Bài giải
a, \(A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{647\cdot650}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{647}-\frac{1}{650}\)
\(A=\frac{1}{5}-\frac{1}{650}=\frac{13}{650}-\frac{1}{650}=\frac{12}{650}=\frac{6}{325}\)
b, \(B=\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{196\cdot200}\)
\(B=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{196\cdot200}\right)\)
\(B=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{196}-\frac{1}{200}\right)\)
\(B=3\left(\frac{1}{3}-\frac{1}{200}\right)=3\cdot\frac{197}{600}=\frac{197}{200}\)
\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-...-\frac{1}{23.27}=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{23.27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\frac{8}{27}=\frac{23}{54}\)
\(A=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{81\cdot85}+\frac{5}{85\cdot89}\\ A=\frac{5}{4}\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{81\cdot85}+\frac{4}{85\cdot89}\right)\\ A=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{81}-\frac{1}{85}+\frac{1}{85}-\frac{1}{89}\right)\\ A=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{89}\right)\\ A=\frac{5}{4}\left(\frac{89}{267}-\frac{3}{267}\right)\\ A=\frac{5}{4}\cdot\frac{86}{267}=\frac{215}{534}\)
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
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)
\(B=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
Vậy giá trị của biểu thức \(B=\frac{32}{99}\)
Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{95}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)