A = \(\dfrac{2}{5.7}\) + \(\dfrac{5}{7.12}\) + \(\dfrac{7}{12.19}\) + \(\dfrac{9}{19.28}\) + \(\dfrac{11}{28.39}\) + \(\dfrac{1}{30.40}\)

A = \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{19}\)  + \(\dfrac{1}{19}\) - \(\dfrac{1}{28}\) + \(\dfrac{1}{28}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)

A = \(\dfrac{1}{5}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)

A = \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\) 

B = \(\dfrac{1}{20}\) + \(\dfrac{1}{44}\) + \(\dfrac{1}{77}\) + \(\dfrac{1}{119}\) + \(\dfrac{1}{170}\)

B = 2 \(\times\) ( \(\dfrac{1}{2.20}\) + \(\dfrac{1}{2.44}\) + \(\dfrac{1}{2.77}\) + \(\dfrac{1}{2.119}\) + \(\dfrac{1}{2.170}\))

B = 2 \(\times\) ( \(\dfrac{1}{40}\) + \(\dfrac{1}{88}\) + \(\dfrac{1}{154}\) + \(\dfrac{1}{238}\) + \(\dfrac{1}{340}\))

B = 2 \(\times\) ( \(\dfrac{1}{5.8}\) + \(\dfrac{1}{8.11}\) + \(\dfrac{1}{11.14}\) + \(\dfrac{1}{14.17}\) + \(\dfrac{1}{17.20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\)+ \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\) + \(\dfrac{3}{17.20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) + \(\dfrac{1}{17}\) - \(\dfrac{1}{20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{20}\))

B = \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{20}\)

B = \(\dfrac{1}{10}\)  = \(\dfrac{34}{340}\) < \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\)

Vậy A > B 

giúp với

\(B=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2021.2014}\)

\(\Rightarrow B=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2021}-\dfrac{1}{2014}\right)\)

\(\Rightarrow B=\dfrac{1}{3}.\left(1-\dfrac{1}{2014}\right)\)

\(\Rightarrow B=\dfrac{1}{3}.\dfrac{2013}{2014}=\dfrac{671}{2014}\)

\(B=\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+...+\dfrac{1}{2021\cdot2024}\\ =\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2021\cdot2024}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2024}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{2024}\right)\\ =\dfrac{1}{3}\cdot\dfrac{2023}{2024}\\ =\dfrac{2023}{6072}\)

\(A=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}+\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}-\dfrac{9}{11}-\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\left(+\dfrac{7}{9}\rightarrow-\dfrac{7}{9}\right)\)

\(\Rightarrow A=\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{5}{7}+\dfrac{7}{9}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}\)

\(\Rightarrow A=-\dfrac{11}{13}+\dfrac{13}{15}\)

\(\Rightarrow A=\dfrac{-11.15+13.13}{13.15}\)

\(\Rightarrow A=\dfrac{-165+169}{195}=\dfrac{4}{195}\)

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`#040911`

\(x-\dfrac{1}{6}=x+\dfrac{5}{7}\\ \Rightarrow x-x=\dfrac{1}{6}+\dfrac{5}{7}\\\Rightarrow0=\dfrac{37}{42}\left(\text{vô lý}\right)\\ \text{Vậy, x không có giá trị nào thỏa mãn.} \)

 2x - 3 = (1/3)*(5 - 4x)
<=>3*(2x - 3) = (5 - 4x)
<=>6x - 9 = 5 - 4x
<=>6x + 4x = 5 + 9
<=>10x = 14
<=>x = 14/10
<=>x = 7/5

`#040911`

\(2x-3=\dfrac{1}{3}\cdot\left(5-4x\right)\)

\(\Leftrightarrow2x-3=\dfrac{5}{3}-\dfrac{4}{3}x\\ \Leftrightarrow2x-3-\dfrac{5}{3}+\dfrac{4}{3}x=0\\ \Leftrightarrow\left(2-\dfrac{4}{3}\right)x+\left(-3-\dfrac{5}{3}\right)=0\\ \Leftrightarrow\dfrac{2}{3}x-\dfrac{14}{3}=0\)

\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{14}{3}\\ \Leftrightarrow x=\dfrac{14}{3}\div\dfrac{2}{3}\\ \Leftrightarrow x=7\\ \text{Vậy, x = 7.}\)

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