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3. + \(20A=\frac{21-1}{1\cdot21}+\frac{22-2}{2\cdot22}+...+\frac{100-80}{80\cdot100}\)
\(\Rightarrow20A=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+...+\frac{1}{80}-\frac{1}{100}\)
\(\Rightarrow20A=\left(1+\frac{1}{2}+...+\frac{1}{80}\right)-\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{1}{20}\left[\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\right]\)
+ \(80B=\frac{81-1}{1\cdot81}+\frac{82-2}{2\cdot82}+...+\frac{100-2}{20\cdot100}\)
\(=1-\frac{1}{81}+\frac{1}{2}-\frac{1}{82}+...+\frac{1}{20}-\frac{1}{100}\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)
\(\Rightarrow B=\frac{1}{80}\left[\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\right]\)
Do đó : \(\frac{A}{B}=\frac{\frac{1}{20}}{\frac{1}{80}}=4\)
4. + \(A=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}=\frac{100}{2}=50\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=3-\left(1-\frac{1}{8}\right)\)
\(A=3-\frac{5}{8}\)
\(A=\frac{19}{8}\)
a) \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
b) \(\frac{1}{q}\left(\frac{1}{n}-\frac{1}{n+q}\right)=\frac{1}{q}\left(\frac{n+q}{n\left(n+q\right)}-\frac{n}{n\left(n+q\right)}\right)=\frac{1}{q}.\frac{q}{n\left(n+q\right)}=\frac{1}{n\left(n+q\right)}\)
a/ Xét mẫu số VP_ n và n+1 là 2 số liên tiếp
\(\Rightarrow\left(n,n+1\right)\)bằng 1
Thay vào đề bài \(\frac{1}{n}-\frac{1}{n+1}\)bằng \(\frac{n+1}{n.\left(n+1\right)}-\frac{n}{n.\left(n+1\right)}\)bằng \(\frac{1}{n\cdot\left(n+1\right)}\)
\(\Rightarrowđpcm\)
P/s _laptop ko gõ đc dấu
a/ \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\)
=> \(A=\frac{9}{10}\)
b/ \(A=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=\frac{n-5}{n-5}+\frac{7}{n-5}\)
=> \(A=1+\frac{7}{n-5}\)
Để A nguyên => 7 chia hết cho n-5 => n-5=(-7; -1; 1; 7)
=> n=(-2; 4, 6, 8)