chứng minh 3/5.11 + 5/11.21 + 7/21.35 + 9/35.53 < 1
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\(\frac{3}{5.11}+\frac{5}{11.21}+\frac{7}{21.35}+\frac{9}{35.53}=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{21}+\frac{1}{21}-\frac{1}{35}+\frac{1}{35}-\frac{1}{53}\right)=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{53}\right)=\frac{1}{10}-\frac{1}{106}
\(\frac{3}{13}.\frac{5}{9}+\frac{1}{6}:\frac{13}{3}+1\)
\(=\frac{3}{13}.\frac{5}{9}+\frac{1}{6}.\frac{3}{13}+1\)
\(=\frac{3}{13}.\left(\frac{5}{9}+\frac{1}{6}\right)+1\)
\(=\frac{3}{13}.\left(\frac{30+9}{54}\right)+1\)
\(=\frac{3}{13}.\frac{39}{54}+1\)
\(=\frac{1}{6}+1\)
\(=\frac{7}{6}\)
\(\frac{5}{6}-\frac{7}{9}.\frac{2}{13}-\frac{7}{9}.\frac{11}{13}+\frac{-2}{9}\)
\(=\frac{5}{6}-\frac{7}{9}.\left(\frac{2}{13}-\frac{11}{13}\right)+\frac{-2}{9}\)
\(=\frac{5}{6}-\frac{7}{9}.\frac{-9}{13}-\frac{2}{9}\)
\(=\frac{5}{6}-\frac{-7}{13}-\frac{2}{9}\)
\(\frac{5}{6}-\frac{7}{9}.\frac{2}{13}-\frac{7}{9}.\frac{11}{13}+\frac{-2}{9}\)
\(=\frac{5}{6}-\frac{7}{9}.\left(\frac{2}{13}-\frac{11}{13}\right)+\frac{-2}{9}\)
\(=\frac{5}{6}-\frac{7}{9}.\frac{-9}{13}-\frac{2}{9}\)
\(=\frac{5}{6}-\frac{-7}{13}-\frac{2}{9}\)
\(=\frac{5}{6}+\frac{7}{13}-\frac{2}{9}\)
\(=\frac{195+126-52}{234}\)
\(=\frac{269}{234}\)
\(\frac{3}{13}.\frac{5}{9}+\frac{1}{6}:\frac{13}{3}+1\)
\(=\frac{3}{13}.\frac{5}{9}+\frac{1}{6}.\frac{3}{13}+1\)
\(=\frac{3}{13}.\left(\frac{5}{9}+\frac{1}{6}\right)+1\)
\(=\frac{3}{13}.\left(\frac{30+9}{54}\right)+1\)
\(=\frac{3}{13}.\frac{39}{54}+1\)
\(=\frac{1}{6}+1=\frac{1}{6}+\frac{6}{6}\)
\(=\frac{7}{6}\)
\(\frac{-7}{9}.\frac{2}{13}-\frac{7}{9}.\frac{11}{13}+\frac{-2}{9}\)
\(=\frac{-7}{9}.\frac{2}{13}+\frac{-7}{9}.\frac{11}{13}+\frac{-2}{9}\)
\(=\frac{-7}{9}.\left(\frac{2}{13}+\frac{11}{13}\right)+\frac{-2}{9}\)
\(=\frac{-7}{9}.1+\frac{-2}{9}\)
\(=\frac{-7}{9}+\frac{-2}{9}\)
\(=\frac{-9}{9}=-1\)
\(\frac{2}{13}.\frac{2}{7}.5\)
\(=\frac{2.2.5}{13.7}\)
\(=\frac{20}{91}\)
\(\frac{1}{5}.\frac{11}{12}.\frac{21}{6}\)
\(=\frac{11.21}{5.12.6}\)
\(=\frac{231}{360}=\frac{77}{120}\)
Giúp mik thêm câu
Tìm các chữ số a,b,c khác 0 thỏa mãn
A)abbc=ab×ac×7
B)tìm n thuộc N sao cho 18n+3 chia hết cho 7
\(S=3^1+3^3+3^5+3^7+...+3^{2015}\)
\(\Rightarrow S=3^1+9.3+9.3^3+.....+9.3^{2013}\)
\(\Rightarrow S=3^1+9.\left(3+3^3+....+3^{2013}\right)\)
Vì \(9.\left(3+3^3+...+3^{2013}\right)⋮9\)mà 31 không chia hết cho 9
=>S không chia hết cho 9
a) 7104 - 1 = (74)26 - 1 = ...1 - 1 = ...0 \(⋮\)5
b) 3201 + 2 = (34)50 . 3 + 2 = ...3 + 2 = ...5 \(⋮\)5
3/5 . 11/7 + 3/5 . (- 4/7) + 2/5
= 3/5 . [11/7 + (- 4/7)] + 2/5
= 3/5 . 1 + 2/5
= 3/5 + 2/5
= 1
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