7\(\frac{3}{10}\)
K
Khách
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D
1
4 tháng 4 2019
1/10 A =7/10^2+7/10^3+..............+7/10^2020
9/10*A=(7/10+7/10^2+......................+7/10^2019)-(7/10^2+7/10^3+........+7/10^2020)
=7/10-7/10^2020
A=10/9 .(7/10-7/10^2020)
NA
0
2 tháng 7 2017
đặt \(A=\frac{7}{10}+\frac{7}{10^2}+\frac{7}{10^3}+\frac{7}{10^4}\)
\(A=7.\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)
Lại đặt \(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\)
\(10B=1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\)
\(10B-B=\left(1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\right)-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)
\(9B=1-\frac{1}{10^4}\)
\(\Rightarrow B=\frac{1-\frac{1}{10^4}}{9}\)
\(\Rightarrow A=7.\frac{1-\frac{1}{10^4}}{9}=\frac{7.\left(1-\frac{1}{10^4}\right)}{9}\)
HD
22 tháng 3 2015
= \(\frac{5\times\left(\frac{1}{7}+\frac{1}{3}-\frac{1}{9}\right)}{10\times\left(\frac{1}{7}+\frac{1}{3}-\frac{1}{9}\right)}\)
=\(\frac{5}{10}\)
=\(\frac{1}{2}\)
CH
Cô Hoàng Huyền
Admin
VIP
2 tháng 3 2018
\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(A=\frac{5.31-\frac{5.2}{7}-\frac{5}{11}+\frac{5}{23}}{13.31-\frac{13.2}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)
\(A=\frac{5.31-\frac{5.2}{7}-\frac{5}{11}+\frac{5}{23}}{13.31-\frac{13.2}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)
\(A=\frac{5}{13}+\frac{1}{3}=\frac{44}{13}\)
2 tháng 3 2018
Bạn tham khảo nhé
Ta có :
\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(A=\frac{5.31-5.\frac{2}{7}-5.\frac{1}{11}+5.\frac{1}{23}}{13.31-13.\frac{2}{7}-13.\frac{1}{11}+13.\frac{1}{23}}+\frac{3.\frac{1}{5}+3.\frac{1}{13}-3.\frac{3}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)
\(A=\frac{5\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}+\frac{3\left(\frac{1}{5}+\frac{1}{13}-\frac{3}{10}\right)}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)
\(A=\frac{5}{13}+\frac{3}{1}=\frac{5}{13}+\frac{39}{13}=\frac{44}{13}\)
Vậy \(A=\frac{44}{13}\)
2 tháng 5 2019
a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)
ta có :
\(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)
\(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)
\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)
Vậy \(A< 3\)
KN
2 tháng 5 2019
a. Ta có :
\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)
\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)
\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)
Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)
Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)
Vậy \(A< 3\)
20 tháng 9 2015
\(C=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(=\frac{5\cdot\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13\cdot\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)
\(=\frac{5}{13}+\frac{3\cdot\left(\frac{1}{5}+\frac{1}{13}-\frac{3}{10}\right)}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)
\(=\frac{5}{13}+3\)
\(=\frac{44}{13}\)
73/10