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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 \(\dfrac{25}{103}< \dfrac{74}{295}\)
\(B=\left(\dfrac{7}{19\cdot31}+\dfrac{5}{19\cdot43}+\dfrac{3}{23\cdot43}+\dfrac{11}{23\cdot57}\right):\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(\Rightarrow\dfrac{1}{2}B=\left(\dfrac{7}{38\cdot31}+\dfrac{5}{38\cdot43}+\dfrac{3}{46\cdot43}+\dfrac{11}{46\cdot57}\right):\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(\Rightarrow\dfrac{1}{2}B=\left(\dfrac{1}{31}-\dfrac{1}{38}+\dfrac{1}{38}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}\right):\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(\Rightarrow\dfrac{1}{2}B=\left(\dfrac{1}{31}-\dfrac{1}{57}\right):\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(\Rightarrow\dfrac{1}{2}B=1\)
\(\Rightarrow B=1:\dfrac{1}{2}=2\)
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
\(S=\dfrac{3}{5.7}+\dfrac{3}{7.9}+....+\dfrac{3}{59.61}\)
\(S=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+......+\dfrac{1}{59}-\dfrac{1}{61}\)
\(S=\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{9}\right)+...+\left(\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(S=\dfrac{1}{5}-\dfrac{1}{61}\)
\(S=\dfrac{56}{305}\)
Vậy S = \(\dfrac{56}{305}\)
\(S=\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
\(S=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(S=\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}.\dfrac{56}{305}=\dfrac{84}{305}\)
