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Gọi \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\)là \(S\)
\(S=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\\ S>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{100\cdot101}\\ S>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ S>\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{5}\)
Vậy \(S>\dfrac{1}{5}\)(đpcm)
\(M=\dfrac{5^3}{1\cdot6}+\dfrac{5^3}{6\cdot11}+...+\dfrac{5^3}{26\cdot31}\)
\(=5^2\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)
\(=5^2\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(=5^2\left(1-\dfrac{1}{31}\right)\)\(=25\cdot\dfrac{30}{31}=\dfrac{750}{31}\)
Câu hỏi của Nguyễn Hải Văn - Toán lớp 6 - Học toán với OnlineMath
a) \(\dfrac{4}{7}=\dfrac{4\cdot9}{7\cdot9}=\dfrac{36}{63}\)
\(\dfrac{13}{9}=\dfrac{13\cdot7}{9\cdot7}=\dfrac{91}{63}\)
\(\dfrac{8}{21}=\dfrac{8\cdot3}{21\cdot3}=\dfrac{24}{63}\)
b) \(\dfrac{1}{-36}=\dfrac{1\cdot5}{-36\cdot5}=\dfrac{-5}{180}\)
\(\dfrac{-8}{45}=\dfrac{-8\cdot4}{45\cdot4}=\dfrac{-32}{180}\)
\(\dfrac{13}{90}=\dfrac{13\cdot2}{90\cdot2}=\dfrac{26}{180}\)
c) \(3=\dfrac{3}{1}=\dfrac{3\cdot23}{1\cdot23}=\dfrac{69}{23}\)
\(-1=\dfrac{-1}{1}=\dfrac{-1\cdot23}{1\cdot23}=\dfrac{-23}{23}\)
\(\dfrac{17}{23}\) giữ nguyên
Link này bạn: Câu hỏi của Quỳnh Anh Shuy - Toán lớp 7 | Học trực tuyến
Đặt A =\(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\)\(-\)\(\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)
A =\(\dfrac{2.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}\)\(-\)\(\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\)
A = \(\dfrac{2}{7}-\dfrac{2}{7}=0\)
~ Chúc bạn học tốt ~
\(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}=\dfrac{2.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{2.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{8}+\dfrac{1}{10}\right)}=1-\dfrac{1}{\dfrac{7}{2}}=1-\dfrac{2}{7}=\dfrac{5}{7}\)