Question

Cho A = \(\dfrac{2}{1.4}\)+\(\dfrac{2}{4.7}\)+\(\dfrac{2}{7.10}\)+.........+\(\dfrac{2}{\left(3x+1\right).\left(3x+4\right)}\)= \(\dfrac{1344}{2017}\)

Chứng tỏ \(\dfrac{7}{12}< A< \dfrac{5}{6}\)

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Answer 1

\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{\left(3x+1\right).\left(3x+4\right)}\)=\(\dfrac{1344}{2017}\)

\(A=\dfrac{2}{3}(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{3x+1}-\dfrac{1}{3x+4}\))=\(\dfrac{1344}{2017}\)

\(A=\dfrac{2}{3}(1-\dfrac{1}{3x+4})\)=\(\dfrac{1344}{2017}\)

\(A=1-\dfrac{1}{3x+4}=\dfrac{1344}{2017}:\dfrac{2}{3}\)

\(A=1-\dfrac{1}{3x+4}=\dfrac{2016}{2017}\)

\(A=\dfrac{1}{3x+4}=1-\dfrac{2016}{2017}\)

\(A=\dfrac{1}{3x+4}=\dfrac{1}{2017}\)

\(\Rightarrow\)\(3x+4=2017\)

\(3x=2017-4\)

\(3x=2013\)

\(x=671\)

\(\Leftrightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)

\(\rightarrowđpcm\)