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A =\(\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
A = \(\dfrac{4}{3}.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{65.68}\right)\)
A = \(\dfrac{4}{3}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\left(\dfrac{1}{11}-\dfrac{1}{11}\right)-...-\left(\dfrac{1}{65}-\dfrac{1}{65}\right)-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-0-0-0-...-0-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\dfrac{33}{68}\)
A = \(\dfrac{11}{17}\)
a, Ta có: \(\dfrac{32}{37}>\dfrac{32}{54}>\dfrac{19}{54}\Rightarrow\dfrac{32}{37}>\dfrac{19}{54}\)
b, Ta có: \(\dfrac{18}{53}>\dfrac{18}{54}=\dfrac{1}{3}\Rightarrow\dfrac{18}{53}>\dfrac{1}{3}\left(1\right)\)
\(\dfrac{26}{78}=\dfrac{1}{3}\left(2\right)\)
Từ (1) và (2) ta suy ra \(\dfrac{18}{53}>\dfrac{26}{78}\)
c, Ta thấy: \(\dfrac{25}{103}< \dfrac{25}{100}=\dfrac{1}{4}\left(1\right)\)
\(\dfrac{74}{295}>\dfrac{74}{296}=\dfrac{1}{4}\left(2\right)\)
Từ (1) và (2) ta suy ra \(\dfrac{25}{103}< \dfrac{74}{295}\)
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
b)\(\dfrac{1}{7}B=\dfrac{1}{10.18}+\dfrac{1}{18.26}+\dfrac{1}{26.34}+...+\dfrac{1}{802.810}\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{8}{10.18}+\dfrac{8}{18.26}+\dfrac{8}{26.34}+...+\dfrac{8}{802.810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{34}+...+\dfrac{1}{802}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}.\dfrac{8}{81}\)
\(\dfrac{1}{7}B=\dfrac{1.8}{8.81}\)
\(\dfrac{1}{7}B=\dfrac{1}{81}\)
\(B=\dfrac{1}{81}:\dfrac{1}{7}\)
\(B=\dfrac{7}{81}\)
Link này bạn: Câu hỏi của Quỳnh Anh Shuy - Toán lớp 7 | Học trực tuyến
Đặt:
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{64}+\dfrac{1}{128}+\dfrac{1}{256}\)
\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}+\dfrac{1}{2^8}\)
\(2A=2\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}+\dfrac{1}{2^8}\right)\)
\(2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}\)
\(2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}+\dfrac{1}{2^8}\right)\)\(A=1-\dfrac{1}{2^8}=1-\dfrac{1}{256}=\dfrac{255}{256}\)