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`@` `\text {Ans}`
`\downarrow`
`1/3+6/9+4/5+5/25`
`= 1/3+2/3+4/5+1/5`
`= 1+1`
`= 2`
`4/7*2/5+4/7*3/5-1/7`
`= 4/7*(2/5+3/5)-1/7`
`= 4/7*1-1/7`
`= 4/7-1/7=3/7`
.....................................................
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\(\frac{9}{11};\frac{7}{9};\frac{5}{7};\frac{3}{5};\frac{1}{3}\)
hok tốt ~
ta thấy : 1 - 1/3 = 2/3 , 1 - 3/5 = 2/5 , 1 - 5/7 = 2/7 , 1-7/9 = 2/9 , 1-9/11 = 2/11
mà : 2/3 > 2/5 > 2/7 > 2/9 > 2/11 nên 1/3 < 1/5 < 1/7 < 1/9 < 1/11
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=17/6:(1-2/3)
=17/6:1/3
=17/2
=13/6×9/2-6/7
=39/4-6/7
=249/28
a) \(\left(\frac{5}{2}+\frac{1}{3}\right):\left(1-\frac{2}{3}\right)=\left(\frac{15}{6}+\frac{2}{6}\right):\frac{1}{3}\)
\(=\frac{17}{6}:\frac{1}{3}=\frac{17}{6}\cdot\frac{3}{1}=\frac{17}{2}\cdot\frac{1}{1}=\frac{17}{2}\)
b) \(\left(\frac{5}{2}-\frac{1}{3}\right)\cdot\frac{9}{2}-\frac{6}{7}=\left(\frac{15}{6}-\frac{2}{6}\right)\cdot\frac{9}{2}-\frac{6}{7}\)
\(=\frac{13}{6}\cdot\frac{9}{2}-\frac{6}{7}=\frac{13}{2}\cdot\frac{3}{2}-\frac{6}{7}=\frac{39}{4}-\frac{6}{7}=\frac{273}{28}-\frac{24}{28}=\frac{249}{28}\)
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A=(6 : 35−116 x 67) : (415 x 1011+5211)A=(6 : 35−116 x 67) : (415 x 1011+5211)
A=(6 : 35−76 x 67) : (215 x 1011+5711)A=(6 : 35−76 x 67) : (215 x 1011+5711)
A=(10−1) : (4211+5711)A=(10−1) : (4211+5711)
A=9 : 9A=9 : 9
A=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Leftrightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
\(A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{51}{3}-\frac{22}{3}=\frac{51-22}{3}=\frac{29}{3}\)
b) \(\frac{5}{12}+\frac{5}{6}-\frac{3}{4}=\frac{5}{12}+\frac{10}{12}-\frac{9}{12}=\frac{5+10-9}{12}=\frac{6}{12}=\frac{1}{2}\)
c) \(1-\left(\frac{1}{5}+\frac{1}{2}\right)=\frac{10}{10}-\frac{2}{10}-\frac{5}{10}=\frac{10-5-2}{10}=\frac{3}{10}\)
d) \(\frac{111}{4}-\left(\frac{25}{7}+\frac{51}{4}\right)=\frac{777}{28}-\frac{60}{28}-\frac{357}{28}=\frac{360}{28}=\frac{90}{7}\)
e) \(\left(\frac{85}{11}+\frac{35}{7}\right)-\frac{35}{11}=\left(\frac{85}{11}-\frac{35}{11}\right)+\frac{35}{7}=\frac{50}{11}-\frac{35}{7}=\frac{350}{77}-\frac{385}{77}=-\frac{35}{77}\)
Ta có : \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+.....+\frac{3}{9.11}\)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{9.11}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{11}\right)=\frac{3}{2}.\frac{10}{11}=\frac{15}{11}\)
Ta có :
\(\frac{3}{1.3}+\frac{3}{3.5}+........+\frac{3}{9.11}\)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+.....+\frac{3}{9.11}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{11}\right)\)
\(=\frac{3}{2}.\frac{10}{11}=\frac{30}{22}\)