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\(A=\dfrac{4}{7.31}+\dfrac{6}{7.41}+\dfrac{9}{10.41}+\dfrac{7}{10.57}\Rightarrow\dfrac{A}{5}=\dfrac{4}{35.31}+\dfrac{6}{35.41}+\dfrac{9}{50.41}+\dfrac{7}{50.57}\)
\(\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\)
\(\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\Rightarrow A=\dfrac{130}{31.57}\)
\(\dfrac{B}{2}=\dfrac{7}{38.31}+\dfrac{5}{38.41}+\dfrac{3}{46.43}+\dfrac{11}{46.57}\Rightarrow\dfrac{B}{2}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\)
\(\Rightarrow B=\dfrac{52}{31.57}\)
\(\dfrac{A}{B}=\dfrac{130}{31.51}:\dfrac{52}{31.57}=\dfrac{5}{2}\)
Ta có : `A:5=4/35.31+6/35.41++9/50.41+7/50.57`
`A/5 = 1/31 - 1/35 +1/35-1/41+1/41-1/50+1/50-1/57`
`A/5 =1/31-1/57` `(1)`
Lại có : `B:2=7/38.31 +5/38.43+3/46.43+11/46.57`
`B/2 =1/31-1/38+1/38-1/43+1/43-1/46+1/46-1/57`
`B/2 =1/31-1/57` `(2)`
Từ `(1)` và `(2)` `=>A/5 =B/2`
`=>A/B=5/2`
\(B=\left(\dfrac{2020}{2}+1\right)+\left(\dfrac{2019}{3}+1\right)+...+\left(\dfrac{1}{2021}+1\right)+1\)
\(=\dfrac{2022}{2}+\dfrac{2022}{3}+...+\dfrac{2022}{2021}+\dfrac{2022}{2022}\)
=2022(1/2+1/3+...+1/2021+1/2022)
=>B/A=2022
Ta có :
B = \(\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)
B = \(\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)
B = \(\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+1\)
B = \(2021\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+...+\dfrac{1}{2}\right)\) (1)
Mà A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{A}{B}=\dfrac{1}{2021}\)
Ta có: \(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)
\(=\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)
\(=\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+\dfrac{2021}{2021}\)
Suy ra: \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{2021\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=\dfrac{1}{2021}\)
Câu 2:
\(\Leftrightarrow x\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{78}\right)=\dfrac{220}{39}\)
\(\Leftrightarrow2x\left(\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{156}\right)=\dfrac{220}{39}\)
\(\Leftrightarrow x\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{12}-\dfrac{1}{13}\right)=\dfrac{110}{39}\)
\(\Leftrightarrow x\cdot\dfrac{10}{39}=\dfrac{110}{39}\)
=>x=11
\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)
\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)
=>B<1
=>A>B
a) \(2\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+...+\dfrac{16}{n\left(n+16\right)}\right)=\dfrac{16}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{n+13}{3\left(n+16\right)}=\dfrac{8}{25}\)
\(24n+384=25n+325\)
\(25n-24n=384-325\)
\(n=59\)
a) Ta có: \(A=\dfrac{4}{7\cdot31}+\dfrac{6}{7\cdot41}+\dfrac{9}{10\cdot41}+\dfrac{7}{10\cdot57}\)
\(=\dfrac{20}{31\cdot35}+\dfrac{30}{35\cdot41}+\dfrac{45}{41\cdot50}+\dfrac{35}{50\cdot57}\)
\(=5\left(\dfrac{4}{31\cdot35}+\dfrac{6}{35\cdot41}+\dfrac{9}{41\cdot50}+\dfrac{7}{50\cdot57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
Ta có: \(B=\dfrac{7}{19\cdot31}+\dfrac{5}{19\cdot43}+\dfrac{3}{23\cdot43}+\dfrac{11}{23\cdot57}\)
\(=\dfrac{14}{31\cdot38}+\dfrac{10}{38\cdot43}+\dfrac{6}{43\cdot46}+\dfrac{22}{46\cdot57}\)
\(=2\left(\dfrac{7}{31\cdot38}+\dfrac{5}{38\cdot43}+\dfrac{3}{43\cdot46}+\dfrac{11}{46\cdot57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{38}+\dfrac{1}{38}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
Suy ra: \(\dfrac{A}{B}=\dfrac{5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}{2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}=\dfrac{5}{2}\)