\(\frac{3}{1x2}+\frac{3}{2x3}+\frac{3}{3x4}+\frac{3}{4x5}+\frac{3}{5x6}+....+\frac{3}{9x10}+\frac{77}{2x9}+\frac{77}{9x16}+\frac{77}{16x23}+...+\frac{77}{93x100}\)
\(\frac{4}{3x6}+\frac{4}{6x9}+\frac{4}{9x12}+\frac{4}{12x15}\) \(\frac{7}{1x5}+\frac{7}{5x9}+\frac{7}{9x13}+\frac{7}{13x17}+\frac{7}{17x21}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\) \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{138}+\frac{1}{340}\)
Toán quá dễ. Tự túc là hạnh phúc mọi nhà bn nhé !
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+\frac{3}{5.6}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\)
Gọi \(\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+......+\frac{3}{9.10}\right)\)là \(A\); \(\left(\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\right)\)là B . Ta có :
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}\cdot\frac{9}{10}=\frac{27}{10}\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{93}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\cdot\frac{49}{100}=\frac{539}{100}\)
\(\Rightarrow\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}=\frac{27}{10}+\frac{539}{100}=\frac{809}{100}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\frac{4}{15}=\frac{16}{45}\)