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1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)
\(B< 1\)
2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)
\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)
\(B=\dfrac{1}{20}\)
3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)
\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)
\(A=11\)
4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)
Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)
\(\Rightarrow B>A\)
=> A=\(\frac{7}{4}\) . ( \(\frac{33}{12}\) + \(\frac{33}{20}\) + \(\frac{33}{30}\) + \(\frac{33}{42}\) ) => A= \(\frac{7}{4}\).33. ( \(\frac{1}{12}\) + \(\frac{1}{20}\) + \(\frac{1}{30}\) + \(\frac{1}{42}\) )
=> A=\(\frac{7}{4}\).33. ( \(\frac{1}{3.4}\) + \(\frac{1}{4.5}\) + \(\frac{1}{5.6}\) + \(\frac{1}{6.7}\) ) = \(\frac{7}{4}\).33.(\(\frac{1}{3}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{6}\) + \(\frac{1}{6}\) - \(\frac{1}{7}\) )
= \(\frac{7}{4}\) .33.(\(\frac{1}{3}\) - \(\frac{1}{7}\)) = \(\frac{7}{4}\) .33. \(\frac{4}{21}\) = 11. Vậy A=11
\(\frac{7}{4}.\left(\frac{101.33}{101.12}+\frac{101.33}{101.20}+\frac{101.33}{101.30}+\frac{101.33}{101.42}\right)\)
\(=\frac{7.33}{4}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\\ =\frac{7.33}{4}\left(\frac{35+21+14+1}{420}\right)\)
\(=\frac{7.3.11}{4}.\frac{71}{420}=\frac{7.3.11.71}{4.4.5.3.7}=\frac{781}{100}\)
mk lm chak vớ vẩn rồi
\(A=\frac{7}{4}\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)
\(A=\frac{7}{4}\left(\frac{385}{140}+\frac{231}{140}+\frac{154}{140}+\frac{110}{140}\right)\)
\(A=\frac{7}{4}.\frac{44}{7}\)
\(A=\frac{44}{4}=11\)
Ta có
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)
\(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\right]\)
\(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)
\(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
\(=\frac{7}{4}.\frac{3333}{101}.\frac{4}{21}=\frac{1111}{101}\)
\(=\dfrac{8}{15}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{42}+\dfrac{33}{56}\right)\)
\(=\dfrac{8}{15}\cdot33\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\right)\)
\(=8\cdot\dfrac{11}{5}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{8}\right)\)
\(=\dfrac{88}{5}\cdot\dfrac{7}{40}=\dfrac{77}{25}\)
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(A=\frac{231}{4}.\frac{4}{21}=\frac{231}{21}=11\)
k nha
\(A=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(A=\frac{7}{4}\left[33\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)
\(A=\frac{7}{4}\left[33\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
\(A=\frac{7}{4}\left[33\times\frac{4}{21}\right]\)
\(A=\frac{7}{4}\times\frac{44}{7}\)
\(A=11\)
A=\(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
A=\(\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
A= \(\frac{7}{4}.\left[33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)
A= \(\frac{7}{4}.\left[33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\right]\)
A= \(\frac{7}{4}.\left[33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)
A= \(\frac{7}{4}.\left[33.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
A= \(\frac{7}{4}.\left[33.\frac{4}{21}\right]\)
A= \(\frac{7}{4}.\frac{44}{7}\)
A= 11
Vậy A= 11
A = 7/4 . (3333/1212 + 3333/2020 + 3333/3030 + 3333/4242)
A = 7/4 . (11/4 + 33/20 + 11/10 + 11/14)
A = 7/4 . 44/7
A = 11
Chúc bạn học tốt
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{42.101}\right)\)
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{42}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{6}\right)\)
\(A=\frac{7}{4}.33.\frac{1}{6}\)
\(A=\frac{7.33}{4.6}\)
\(A=\frac{7.3.11}{4.3.2}\)
\(A=\frac{7.11}{4.2}\)
\(A=\frac{77}{8}\)
Ta có : A = \(\dfrac{7}{4}.\left(\dfrac{333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(\Rightarrow A=\dfrac{7}{4}.\left(\dfrac{11}{4}+\dfrac{33}{20}+\dfrac{11}{10}+\dfrac{11}{14}\right)\)
\(\Rightarrow A=\dfrac{7}{4}.\dfrac{44}{7}\)
\(\Rightarrow A=11\)
Vậy A = 11
A=7/4(33/12+33/20+33/30+33/42)
=7/4.33.(1/12+1/20+1/30+1/42)
=7/4.33.4/21
=11