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\(\Rightarrow A=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{59.61}\right)\)
\(\Rightarrow A=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{59}-\frac{1}{61}\right)\)
\(\Rightarrow A=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(\Rightarrow A=\frac{3}{2}.\frac{56}{305}\)
\(\Rightarrow A=\frac{84}{305}\)
\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{19.21}\)
\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{19.21}\right)\)
\(=2\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(=2\left(1-\dfrac{1}{21}\right)=2.\dfrac{20}{21}=\dfrac{40}{21}\)
= \(\frac{2.2}{1.3}+\frac{3.3}{2.4}+\frac{4.4}{3.5}+\frac{5.5}{4.6}+\frac{6.6}{5.7}\)
= \(\frac{2.3.4.5.6}{1.2.3.4.5}+\frac{2.3.4.5.6}{3.4.5.6.7}\)
= \(\frac{2}{1}+\frac{6}{7}\)
= 2\(\frac{6}{7}\)
Mình nghĩ zậy !!!!!!!!!!!!!!!!!!
\(3\frac{1}{4}=\frac{13}{4}=2,25;\)
\(2\frac{3}{8}=\frac{19}{8}=2,375\);
\(5\frac{7}{25}=\frac{132}{25}=5,28\).
Ta có:
\(S=\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{59.61}\)
\(=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=2.\left(\frac{61}{305}-\frac{5}{305}\right)\)
\(=2.\frac{56}{305}\)
\(=\frac{112}{305}\)
Vậy \(S=\frac{112}{305}\)