Tìm x biết 7/4*(3333/1212+3333/2020+3333/3030+3333/4242+3333/5656)*x=-385/32
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DP
10 tháng 7 2017
\(M=\frac{7}{4}\times\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}+\frac{3333}{5656}\right)\)
\(M=\frac{7}{4}\times\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}+\frac{33}{56}\right)\)
\(M=\frac{7}{4}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}+\frac{33}{56}\right)\)
\(M=\frac{7}{4}\times\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}+\frac{33}{7.8}\right)\)
\(M=\frac{7}{4}\times\left[33\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\right]\)
\(M=\frac{7}{4}\times\left[33\times\left(\frac{1}{3}-\frac{1}{8}\right)\right]\)
\(M=\frac{7}{4}\times\left(33\times\frac{5}{24}\right)=\frac{7}{4}\times\frac{55}{8}=\frac{385}{32}\)
PN
1
23 tháng 4 2016
A=7/4.(3333/1212+3333/2020+3333/3030+3333/4242)
A=7/4 X 44/7
A=11
HB
1
NT
5
22 tháng 4 2016
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(\Leftrightarrow A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(\Leftrightarrow A=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(\Leftrightarrow A=\frac{7}{4}.\left[33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\right]\)
\(\Leftrightarrow A=\frac{7}{4}.\left[33.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
\(\Leftrightarrow A=\frac{7}{4}.\left[33.\frac{4}{21}\right]\)
\(\Leftrightarrow A=\frac{7}{4}.\frac{44}{7}\)
\(\Leftrightarrow A=11\)
TH
2
21 tháng 5 2015
A=7/4.(333/1212+3333/2020+3333/3030+3333/4242)
=7/4.3333.(1/1212+1/2020+1/4242)
=7/4.3333.(1/12.101+1/20.101+1/42.101)
=7/4.3333.1/101.(1/12+1/20+1/42)
=7/4.33.101.1/101.(1/3.4+1/4.5+1/5.6+1/6.7)
=7/4.33.(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7)
=7/4.33.(1/3-1/7)
=7/4.33.4/21
=7.33.4/4.21
=11
HL
10
12 tháng 6 2015
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 \(A=33\)
20 tháng 11 2020
\(\frac{7}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\right)\)
\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{7}{4}\cdot33\cdot\frac{4}{21}\)
\(=\left(\frac{7}{4}\cdot\frac{4}{21}\right)\cdot33\)
\(=\frac{1}{3}\cdot33=11\)
P
1
28 tháng 8 2016
\(\frac{7}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)=\frac{7}{4}\left(\frac{33\times101}{12\times101}+\frac{33\times101}{20\times101}+\frac{33\times101}{30\times101}+\frac{33\times101}{42\times101}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=\frac{7}{4}\times\frac{44}{7}=11\)