\(\frac{1}{7}+\frac{?}{?}=1\)
K
Khách
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N
20 tháng 6 2017
\(\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4+\frac{4}{7}+\frac{4}{7^2}-\frac{4}{7^3}}.\frac{858585}{313131}.\left(-1\frac{14}{17}\right)\)\
\(=\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4.\left(1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}\right)}.\frac{85}{31}.\left(-\frac{31}{17}\right)\)
\(=\frac{1}{4}.\frac{85}{31}.\left(-\frac{31}{17}\right)\)
\(=-\frac{5}{4}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
20 tháng 6 2017
\(=\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4\left(1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}\right)}.\frac{85.10101}{31.10101}.\left(-\frac{31}{17}\right).\)
\(=\frac{1}{4}.\frac{3.17}{31}.\left(-\frac{31}{17}\right)=-\frac{3}{4}\)
DP
1 tháng 4 2017
\(B=\frac{7}{3.13}+\frac{7}{13.23}+...+\frac{7}{53.63}\)
\(B=10.\left(\frac{1}{3.13}+\frac{1}{13.23}+....+\frac{1}{53.63}\right)\)
\(B=10.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+...+\frac{1}{53}+\frac{1}{63}\right)\)
\(B=10.\left(\frac{1}{3}-\frac{1}{63}\right)\)
\(B=10.\frac{20}{63}\)
\(B=\frac{200}{63}\)
TM
21 tháng 9 2017
\(P=\frac{\frac{3}{7}-\frac{3}{13}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}\)
\(=\frac{16}{35}\)
17 tháng 9 2015
\(A=\frac{3.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{1.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}{\left(-7\right).\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(A=\frac{3}{5}+\frac{-1}{7}\)
\(A=\frac{21}{35}+\frac{-5}{35}\)
\(A=\frac{16}{35}\)
MH
17 tháng 9 2015
\(A=\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-\frac{7}{2}+\frac{7}{3}-\frac{7}{4}+\frac{7}{5}}\)
\(=\frac{3}{5}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
=3/5+(1/-7)
=3/5-1/7
=16/35
9 tháng 8 2017
a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)
b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0
9 tháng 8 2017
a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}\)
\(=\frac{25}{33}\)
b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)
Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.
DT
22 tháng 1 2019
A= 7/8:(4/18-1/18)+7/8:(1/36-15/36)
=7/8:1/6+7/8:(-7/18)
=7/8:(1/6+-7/18)=7/8:(3/18+-7/18)=7/8:(-2/9)=-63/18=-7/2
18 tháng 3 2018
Ta có :
\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\)\(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\)\(\frac{2}{7}:\frac{2}{7}\)
\(=\)\(\frac{2}{7}.\frac{7}{2}\)
\(=\)\(1\)
Chúc bạn học tốt ~
NY
18 tháng 3 2018
\(=\frac{2-2+2}{7-7+7}:\frac{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2}{7}:\frac{2-2+2}{7-7+7}\)
\(=\frac{2}{7}:\frac{2}{7}\)
\(=\frac{2}{7}.\frac{7}{2}\)
\(=\frac{2.7}{7.2}\)
\(=\frac{1.1}{1.1}\)
\(=\frac{1}{1}\)
\(=1\)
Kết quả là \(\frac{6}{7}\)
1/7+6/7=1