Tính giá trị biểu thức sau một cách hợp lí: \(\dfrac{8}{15}\).\([\)\(\dfrac{3333}{1212}\)+\(\dfrac{3333}{2020}\)+\(\dfrac{3333}{4242}\)+\(\dfrac{3333}{5656}\)\(]\)
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Những câu hỏi liên quan
26 tháng 5 2017
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
26 tháng 5 2017
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
18 tháng 3 2018
\(\frac{8}{15}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{4242}+\frac{3333}{5656}\right)\)
= \(\frac{8}{15}.\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{14}+\frac{33}{56}\right)\)
= \(\frac{8}{15}.\frac{231}{40}\)
= \(\frac{77}{25}\)
NV
5
26 tháng 7 2017
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(A=\frac{231}{4}.\frac{4}{21}\)
\(A=11\)
P
1
28 tháng 8 2016
\(\frac{7}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)=\frac{7}{4}\left(\frac{33\times101}{12\times101}+\frac{33\times101}{20\times101}+\frac{33\times101}{30\times101}+\frac{33\times101}{42\times101}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=\frac{7}{4}\times\frac{44}{7}=11\)
KT
3
19 tháng 4 2016
A= 7/4 (33/12 +33/20+33/30+33/42)
A= 7/4 * 33 * (1/12+1/20+1/30+1/42)
A= 7/4 * 33 * (1/3*4 + 1/4*5 +1/5*6 +1/6*7)
A= 231/4 (1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7)
A=231/4(1/3-1/7)
A=231/4(7/21-3/21)
A=231/4 * 4/21
A=11
Vậy A=11
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.
KS
0
HD
1
7 tháng 5 2021
\(A=\frac{7}{4}\cdot\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}\cdot\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}\cdot33\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
Rồi tách ra rồi làm nốt
\(=\dfrac{8}{15}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{42}+\dfrac{33}{56}\right)\)
\(=\dfrac{8}{15}\cdot33\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\right)\)
\(=8\cdot\dfrac{11}{5}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{8}\right)\)
\(=\dfrac{88}{5}\cdot\dfrac{7}{40}=\dfrac{77}{25}\)