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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 \(A=33\)
DT
1
20 tháng 4 2016
A = 7/4 * ( 3333/1212 + 3333/2020 + 3333/3030 + 3333/4242) => A = 7/4* (33/12 + 33/20 + 33/30 + 33/42) => A = 7/4* ( 33/3*4 + 33/4*5 + 33/5*6 + 33/6*7) => A = 7/4* { 33/(4-3) * ( 1/4 - 1/5 + 1/5 - 1/6 + 1/6-1/7)} => A = 7/4*33 * ( 1/4 - 1/7) A = 231/4 * 3/28 =693/112.
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
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 : \(A=11\)
AC
1
8 tháng 5 2017
A = 7/4.[3333.(1/1212+1/2020+1/3030+1/4242)]
A= 7/4.[3333.(1/12.101+1/20.101+1/30.101+1/42.101)]
A= 7/4.[3333.(1/12+1/20+1/30+1/42)]
A= 7/4.[3333.(1/3.4+1/4.5+1/5.6+1/6.7)]
A= 7/4.[3333.(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7)]
A= 7/4.[3333.(1/3-1/7)]
A= 7/4.[3333.4/21]
A= 7/4.4444/7
A=1111
P
1
28 tháng 8 2016
\(\frac{7}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)=\frac{7}{4}\left(\frac{33\times101}{12\times101}+\frac{33\times101}{20\times101}+\frac{33\times101}{30\times101}+\frac{33\times101}{42\times101}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=\frac{7}{4}\times\frac{44}{7}=11\)
MA
Tính bằng cách hợp lí
H=7/4x(3333/1212+3333/2020+3333/3030+3333/4242)
mình đang cần gấp, giải nhanh hộ
3
3 tháng 7 2018
\(H=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(H=\frac{7}{4}.3333.\left(\frac{1}{1212}+\frac{1}{2020}+\frac{1}{3030}+\frac{1}{4242}\right)\)
\(H=\frac{7}{4}.3333.\left(\frac{1}{12.101}+\frac{1}{20.101}+\frac{1}{30.101}+\frac{1}{42.101}\right)\)
\(H=\frac{7}{4}.3333.\frac{1}{101}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(H=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(H=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(H=\frac{7}{4}.33.\left(\frac{7}{21}-\frac{3}{21}\right)\)
\(H=33.\frac{7}{4}.\frac{4}{21}\)
\(H=11.3.\frac{1}{3}\)
\(H=11\)
Tham khảo nhé~
3 tháng 7 2018
\(H=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\))
\(H=\frac{7}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{30.101}+\frac{33.101}{42.101}\right)\)
\(H=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(H=\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(H=\frac{231}{4}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(H=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(H=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(H=\frac{231}{4}.\frac{4}{21}\)
\(H=11\)
NV
5
26 tháng 7 2017
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(A=\frac{231}{4}.\frac{4}{21}\)
\(A=11\)
\(A=\frac{7}{4}\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)
\(A=\frac{7}{4}.\frac{44}{7}=11\)
chuẩn ròi