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Nhân cả tử cả mẫu của các phân số trong A với 2 ta có:
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..........+\frac{2}{90}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.........+\frac{1}{90}\right)\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{9.10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=2.\frac{2}{5}\)
\(=\frac{4}{5}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{2}{5}\)
\(E=\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{6}-1\right)\cdot\left(\dfrac{1}{10}-1\right)\cdot...\cdot\left(\dfrac{1}{45}-1\right)\)
\(E=1\cdot\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{45}\right)\)
\(E=1\cdot\left(\dfrac{1}{45}-\dfrac{1}{3}\right)\div2+1\)
\(E=\dfrac{1}{22}\)
\(E=\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{6}-1\right)\cdot\left(\dfrac{1}{10}-1\right)\cdot...\cdot\left(\dfrac{1}{45}-1\right)\)
\(\text{E=}\left(\dfrac{1}{3+6+10+...+45}\cdot1\right)\)
\(\text{E = }\left(\dfrac{1}{\left(45-3\right)\div2+1}\right)\)
\(\text{E = }\dfrac{1}{22}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)
= \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
= \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{10}\right)\)
= \(2.\frac{2}{5}\)
= \(\frac{4}{5}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\\ =\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\\ =2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(2\left(\frac{1}{2}-\frac{1}{10}\right)=\frac{4}{5}\)
Ta có :
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}+\frac{1}{50}\)
\(\Rightarrow A=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\right)\)
\(\Rightarrow A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)
\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(\Rightarrow A=2.\left(\frac{11}{22}-\frac{2}{22}\right)\)
\(\Rightarrow A=2.\frac{9}{22}\)
\(\Rightarrow A=\frac{9}{11}\)
a:2=1/6+1/12+1/20+......+1/90+1/110
a:2=1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10+1/10-1/11
a:2 =1/2-1/11 ok nhé