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\(=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\frac{56}{305}\)
\(=\frac{84}{305}\)
tk cho minh nhe >.<
Ta có: \(\frac{2}{3}\times\left(\frac{3}{5.7}+\frac{3}{7.9}+.....+\frac{3}{59.61}\right):\frac{2}{3}\)
\(\Rightarrow\left(\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{59.61}\right):\frac{2}{3}\)
\(=\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\right):\frac{2}{3}\)
\(\Rightarrow\left(\frac{1}{5}-\frac{1}{61}\right):\frac{2}{3}=\frac{56}{305}:\frac{2}{3}=\frac{84}{305}\)
\(=\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{59.61}\)
\(=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
=\(2.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=2.\left(\frac{36}{505}\right)\)
\(=\frac{72}{505}\)
TK nha !!
Ta có : \(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+....+\frac{4}{59.61}\)
\(=2\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+.....+\frac{2}{59.61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=2.\frac{56}{305}=\frac{112}{305}\)
\(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{59.61}\)
\(=\)\(\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{59.61}\right)\)
\(=\)\(\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\)\(\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\)\(\frac{3}{2}.\frac{56}{305}\)
\(=\)\(\frac{84}{305}\)
Chúc bạn học tốt ~
A = 2/3*5 + 2/5*7 + 2/7*9 + ... + 2/97*99
A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99
A = 1/3 - 1/99
A = 32/99
\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(A=\frac{1}{3}-\frac{1}{99}\)
\(A=\frac{32}{99}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
tớ ko chép lại đề đâu
\(\frac{1}{2}M=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{101.103}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{101}-\frac{1}{103}\)
\(=\frac{1}{5}-\frac{1}{103}\)
=\(\frac{98}{515}\)
=> \(M=\frac{98}{515}:\frac{1}{2}=\frac{196}{515}\)
Vậy \(M=\frac{196}{515}\)
\(\frac{6^2}{5\cdot7}\cdot\frac{7^2}{6\cdot8}\cdot\frac{8^2}{7\cdot9}\cdot\frac{9^2}{8\cdot10}\)
=\(\frac{6\cdot6}{5\cdot7}\cdot\frac{7\cdot7}{6\cdot8}\cdot\frac{8\cdot8}{7\cdot9}\cdot\frac{9\cdot9}{8\cdot10}\)
=\(\frac{6\cdot7\cdot8\cdot9}{5\cdot6\cdot7\cdot8}\cdot\frac{6\cdot7\cdot8\cdot9}{7\cdot8\cdot9\cdot10}\)
=\(\frac{9}{5}\cdot\frac{3}{5}\)=\(\frac{27}{25}\)
**** MIK
\(=\frac{6.6}{5.7}.\frac{7.7}{6.8}.\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{6.6.7.7.8.8.9.9}{5.6.7.8.9.10.8.7}\)
\(=\frac{27}{25}\)
khoan đã bạn chép nhầm đề rồi thì phải số 1 kia không có dấu gì à?
Đặt \(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
\(\Rightarrow A=\frac{98}{303}\div2=\frac{49}{303}\)
\(\frac{5}{5.7}+\frac{5}{7.9}+...+\frac{5}{59.61}\)
\(=\frac{5}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{5}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\frac{5}{2}\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{5}{2}.\frac{56}{305}=\frac{28}{61}\)