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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 \(A=33\)
LH
3
TK
1
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{30.101}+\frac{33.101}{42.101}\right)\)
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{7.11}{4}.\left(\frac{1}{4}+\frac{3}{20}+\frac{1}{10}+\frac{1}{14}\right)\)
\(A=\frac{77}{4}.\left(\frac{35}{140}+\frac{21}{140}+\frac{14}{140}+\frac{10}{140}\right)\)
\(A=\frac{77}{4}.\frac{80}{140}=\frac{77}{8}.\frac{20}{35}\)
\(A=11\)
Vậy : \(A=11\)
NQ
2
CM
16 tháng 11 2017
a)
5 7 ⋅ 10 11 + 5 7 ⋅ 14 11 − 5 7 ⋅ 17 11 = 5 7 10 11 + 14 11 − 17 11 = 5 7 .1 = 5 7
b)
1 3 ⋅ 4 5 + 1 3 ⋅ 1 1 5 − − 2 3 1 = 1 3 ⋅ 4 5 + 1 3 ⋅ 6 5 + 2 3 = 1 3 4 5 + 6 5 + 2 3 = 1 3 .2 + 2 3 = 4 3
PJ
1
23 tháng 7 2018
2323/202+2323/606+2323/1212+2323/2020+2323/3030+2323/4242+2323/5656+2323/7272+2323/9090.=
\(\frac{23\times101}{2\times101}+\frac{23\times101}{6\times101}+\frac{23\times101}{12\times101}+....+\frac{23\times101}{72\times101}+\frac{23\times101}{90\times101} \)
=\(\frac{23}{2}+\frac{23}{6}+..........+\frac{23}{72}+\frac{23}{90}\)
=\(23\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+.....+\frac{1}{9\times10}\right)\)
=\(23\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}_{ }\right)\)
=\(23\left(1-\frac{1}{10}\right)_{ }\)
=\(23\times\frac{9}{10}\)
==\(\frac{207}{10}\)
chúc học giỏi nha bn
\(\frac{2020.2020+3030}{2021.2021-1011}\)
\(=\frac{2020.2020+3030}{\left(2020+1\right)\left(2020+1\right)+3030}\)
\(=\frac{2020.2020+3030}{\left(2020+1\right)2020+2020+1-1011}\)
\(=\frac{2020.2020+3030}{2020.2020+2020+2020+1-1011}\)
\(=\frac{2020.2020+3030}{2020.2020+3030}=1\)