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\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{4}\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{4}\)
\(=\left(\dfrac{1}{2}+\dfrac{1}{4}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\right)\)
\(=\dfrac{3}{4}+\dfrac{1}{2}+1\)
\(=\dfrac{9}{4}\)
\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{4}\)
=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{4}\)
=\(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{4}+\dfrac{1}{3}\right)+\left(\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\right)\)
=\(\dfrac{5}{4}+1\)
=\(\dfrac{9}{4}\)
Ta có : A=\(\dfrac{0,5+\dfrac{7}{12}-\dfrac{5}{6}}{1-\dfrac{2}{3}+0,75}=\dfrac{\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{5}{6}}{1-\dfrac{2}{3}+\dfrac{3}{4}}=\dfrac{\dfrac{1}{4}}{\dfrac{13}{12}}=\dfrac{13}{48}\)
\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right)\cdot\dfrac{-1}{3}+\dfrac{1}{3}+\dfrac{1}{6}:\left(-2\right)\)
\(=\dfrac{-11}{10}\cdot\dfrac{-1}{3}+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)
\(K=5\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{45\cdot46}\right)\)
\(=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{45}-\dfrac{1}{46}\right)\)
=5*45/46=225/46
\(T=\dfrac{1}{5}\cdot\sqrt{6\cdot\dfrac{2}{3}}-\dfrac{3}{2}\cdot\sqrt{\dfrac{2}{3}\cdot\dfrac{8}{75}}+\dfrac{1}{2}\cdot\sqrt{6\cdot\dfrac{8}{75}}\)
\(=\dfrac{1}{5}\cdot2-\dfrac{3}{2}\cdot\dfrac{4}{15}+\dfrac{1}{2}\cdot\dfrac{4}{5}\)
=2/5-12/30+4/10
=2/5
a: \(=0.5\cdot10-\dfrac{1}{7}+15=20-\dfrac{1}{7}=\dfrac{139}{7}\)
b: \(=6\cdot\dfrac{-2}{3}+12\cdot\dfrac{4}{9}+18\cdot\dfrac{-8}{27}\)
\(=-4+\dfrac{16}{3}-\dfrac{16}{3}=-4\)
c: \(=\left(\dfrac{5}{2}+\dfrac{3}{8}-\dfrac{5}{8}+\dfrac{2}{3}\right):\left(\dfrac{17}{2}+\dfrac{49}{4}-\dfrac{17}{8}+\dfrac{34}{15}\right)\)
\(=\dfrac{35}{12}:\dfrac{2507}{120}=\dfrac{350}{2507}\)
(\(1\dfrac{1}{2}\)−0,5):(−3)\(^2\)+\(\dfrac{1}{3}-\dfrac{1}{6}\)
= 1:9+\(\dfrac{1}{3}-\dfrac{1}{6}\)
= \(\dfrac{1}{9}+\dfrac{1}{3}-\dfrac{1}{6}\)
= \(\dfrac{4}{9}-\dfrac{1}{6}\)
= \(\dfrac{5}{18}\)
\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}=\dfrac{-3\left(-0,125+0,1-\dfrac{1}{11}-\dfrac{1}{12}\right)}{5\left(-0,125+0,1-\dfrac{1}{11}-\dfrac{1}{12}\right)}=\dfrac{-3}{5}\)
A=\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\)
A=\(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{-5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}\)
A=\(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)
A=\(\dfrac{3}{-5}=\dfrac{-3}{5}\)
=1/8-1/5+1/7*3/8-3/5+3/7 + 1/2+1/3-1/5*3/4+1/2-3/10
=19/280*57/280+19/30*19/20
=19/280.280/57+19/30.20/19
=1/1.1/3+1/3.2/1
=1/3+2/3=3/3
=1
Ta có:
\(A=\dfrac{0,5}{3}+\dfrac{0,5}{6}+\dfrac{0,5}{10}+...+\dfrac{0,5}{1275}+\dfrac{0,5}{1326}\)
\(\Rightarrow A=0,5\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{1}{1275}+\dfrac{1}{1326}\right)\)
\(\Rightarrow A=0,5.2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{10}+....+\dfrac{1}{2550}+\dfrac{1}{2652}\right)\)
\(\Rightarrow A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{10}+....+\dfrac{1}{2550}+\dfrac{1}{2652}\)
\(\Rightarrow A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{50.51}+\dfrac{1}{51.52}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{50}-\dfrac{1}{51}+\dfrac{1}{51}-\dfrac{1}{52}\)\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{26}{52}-\dfrac{1}{52}=\dfrac{25}{52}\)