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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{4}{5}\cdot\dfrac{5}{8}:\dfrac{4}{5}=\left(\dfrac{4}{5}:\dfrac{4}{5}\right)\cdot\dfrac{5}{8}=1\cdot\dfrac{5}{8}=\dfrac{5}{8}\)
b) \(\dfrac{5}{6}+\left(\dfrac{1}{2}:\dfrac{3}{2}+\dfrac{4}{5}\right)=\dfrac{5}{6}+\left(\dfrac{1}{2}\cdot\dfrac{2}{3}+\dfrac{4}{5}\right)=\dfrac{5}{6}+\left(\dfrac{1}{3}+\dfrac{4}{5}\right)=\dfrac{17}{15}+\dfrac{5}{6}=\dfrac{59}{30}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(1-\left(5\dfrac{4}{9}+a-7\dfrac{7}{18}\right):15\dfrac{3}{4}=0\)
=>\(\left(5+\dfrac{4}{9}+a-7-\dfrac{7}{18}\right):\dfrac{63}{4}=1\)
=>\(\left(a-2+\dfrac{1}{18}\right)=\dfrac{63}{4}\)
=>\(a-\dfrac{35}{18}=\dfrac{63}{4}\)
=>\(a=\dfrac{63}{4}+\dfrac{35}{18}=\dfrac{637}{36}\)
b: \(B=\left(\dfrac{2}{15}+\dfrac{5}{3}-\dfrac{3}{5}\right):\left(4\dfrac{2}{3}-2\dfrac{1}{2}\right)\)
\(=\dfrac{2+5\cdot5-3^2}{15}:\left(4+\dfrac{2}{3}-2-\dfrac{1}{2}\right)\)
\(=\dfrac{2+4^2}{15}:\left(2+\dfrac{2}{3}-\dfrac{1}{2}\right)\)
\(=\dfrac{18}{15}:\dfrac{13}{6}=\dfrac{6}{5}\cdot\dfrac{6}{13}=\dfrac{36}{65}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
` a/`
` 2 - 1 5/6 + 2 2/3 = 2 - 11/6 - 8/3 = 1/6+ 8/3 = 1/6 + 16/6 = 17/6 `
`b/`
`5/9 xx ( 2 5/6 - 1 2/3 ) = 5/9 xx ( 17/6 - 5/3 ) = 5/9 xx 7/6 = 35/54 `
`c/`
` 1 1/3 : ( 2 + 1 1/6 : 2 5/6 ) `
`= 4/3 : ( 2 + 7/6 : 17/6 ) `
`= 4/3 : ( 2 + 7/6 xx 6/17 )`
`= 4/3 : ( 2 + 7/17 ) `
`= 4/3 : ( 34/17 + 7/17 ) `
`= 4/3 : 41/17 `
`= 4/3 xx 17/41 `
`= 68/123`
` d/`
` 2 3/5 : 3/4 xx 1 4/5 = 13/5 xx 4/3 xx 9/5 =52/15 xx 9/5 = 156/25`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}+\dfrac{4}{5}-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\left(-\dfrac{4}{5}+\dfrac{4}{5}\right)+\left(\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}\right)\)
\(=0+0+0+0-\dfrac{257}{210}\)
\(=\dfrac{257}{210}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 4:
\(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)
\(\Rightarrow y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)
\(\Rightarrow y:\dfrac{2}{5}=\dfrac{25}{16}\)
\(\Rightarrow y=\dfrac{2}{5}\cdot\dfrac{25}{16}\)
\(\Rightarrow y=\dfrac{5}{8}\)
________________
\(456+y:87=23987\)
\(\Rightarrow y:87=23987-456\)
\(\Rightarrow y:87=23531\)
\(\Rightarrow y=23531\cdot87\)
\(\Rightarrow y=2047197\)
a)\(\dfrac{4}{5}\times\dfrac{5}{8}:\dfrac{4}{5}\)
\(=\left(\dfrac{4}{5}:\dfrac{4}{5}\right)\times\dfrac{5}{8}\)
\(=1\times\dfrac{5}{8}=\dfrac{5}{8}\)
b)\(\dfrac{5}{6}+\left(\dfrac{1}{2}:\dfrac{3}{2}+\dfrac{4}{5}\right)\)
\(=\dfrac{5}{6}+\left(\dfrac{1}{3}+\dfrac{4}{5}\right)\)
\(=\dfrac{5}{6}+\dfrac{17}{15}\)
\(=\dfrac{59}{30}\)
Bài 2:
a) \(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)
\(y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)
\(y:\dfrac{2}{5}=\dfrac{25}{16}\)
\(y=\dfrac{25}{16}\times\dfrac{2}{5}\)
\(y=\dfrac{5}{8}\)
b)\(456+y:87=23987\)
\(y:87=23987-456\)
\(y:87=23531\)
\(y=23531\times87\)
\(y=2047197\)
![](https://rs.olm.vn/images/avt/0.png?1311)
TL:
a) \(\dfrac{5}{6}\);\(\dfrac{9}{8}\);\(\dfrac{17}{18}\)
b) \(\dfrac{1}{2}\);\(\dfrac{5}{8}\);\(\dfrac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
1.
$3\frac{1}{4}-2\frac{1}{3}=3+\frac{1}{4}-(2+\frac{1}{3})$
$=3-2+\frac{1}{4}-\frac{1}{3}$
$=1+\frac{1}{4}-\frac{1}{3}=\frac{5}{4}-\frac{1}{3}=\frac{11}{12}$
2.
$6\frac{5}{9}+2\frac{5}{6}=6+\frac{5}{9}+2+\frac{5}{6}=8+\frac{5}{6}+\frac{5}{9}=8+\frac{25}{18}=8+1+\frac{7}{18}=9+\frac{7}{18}=9\frac{7}{18}$
3.
$6\frac{5}{9}-2\frac{5}{6}=6+\frac{5}{9}-(2+\frac{5}{6})$
$=6+\frac{5}{9}-2-\frac{5}{6}$
$=(6-2)+\frac{5}{9}-\frac{5}{6}$
$=4+\frac{5}{9}-\frac{5}{6}=\frac{41}{9}-\frac{5}{6}=\frac{67}{8}$
\(B=\dfrac{5}{3}x\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}\right)=\)
\(=\dfrac{5}{3}x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{9x10}\right)\)
\(C=\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{10-9}{9x10}=\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(\Rightarrow B=\dfrac{5}{3}xC=\dfrac{5}{3}x\dfrac{9}{10}=\dfrac{3}{2}\)