3333 +3333+3333+3333 1212 3030 2020 4242
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\(\frac{7}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\right)\)
\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{7}{4}\cdot33\cdot\frac{4}{21}\)
\(=\left(\frac{7}{4}\cdot\frac{4}{21}\right)\cdot33\)
\(=\frac{1}{3}\cdot33=11\)
A=7/4.(3333/1212+3333/2020+3333/3030+3333/4242)
A=7/4 X 44/7
A=11
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(\Leftrightarrow A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(\Leftrightarrow A=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(\Leftrightarrow A=\frac{7}{4}.\left[33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\right]\)
\(\Leftrightarrow A=\frac{7}{4}.\left[33.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
\(\Leftrightarrow A=\frac{7}{4}.\left[33.\frac{4}{21}\right]\)
\(\Leftrightarrow A=\frac{7}{4}.\frac{44}{7}\)
\(\Leftrightarrow A=11\)
Ta có:
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(=\frac{7}{4}.\left[3333.\left(\frac{1}{1212}+\frac{1}{2020}+\frac{1}{3030}+\frac{1}{4242}\right)\right]\)
\(=\frac{7}{4}.\left[3333.\left(\frac{1}{12.101}+\frac{1}{20.101}+\frac{1}{30.101}+\frac{1}{42.101}\right)\right]\)
\(\frac{7}{4}.\left[3333.\frac{1}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)
\(=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)
\(=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
\(=33.\left(\frac{7}{4}.\frac{4}{7}\right)\)
\(=33.1\)
\(=33\)
Vậy \(A=33\)
\(A=\frac{7}{4}\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)
\(A=\frac{7}{4}.\frac{44}{7}=11\)
\(\frac{7}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)=\frac{7}{4}\left(\frac{33\times101}{12\times101}+\frac{33\times101}{20\times101}+\frac{33\times101}{30\times101}+\frac{33\times101}{42\times101}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=\frac{7}{4}\times\frac{44}{7}=11\)
44/7