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a: \(\Leftrightarrow\dfrac{32}{x}=\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{99}\)
=>32/x=1/3-1/5+1/5-1/7+...+1/9-1/11
=>32/x=1/3-1/11=8/33
=>x=32:8/33=132
b: \(\Leftrightarrow1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{56}=\dfrac{x}{16}\)
\(\Leftrightarrow6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{x}{16}\)
=>x/16=6-1/2+1/8=11/2+1/8=45/8=90/16
=>x=90
c: \(\Leftrightarrow\dfrac{22}{x}=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\)
=>22/x=1/2*2/3*...*9/10*3/2*4/3*...*11/10
=>22/x=1/10*11/2=11/20=22/40
=>x=40
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)4/5+5 1/2 xx (4,5-2)+7/10`
`=4/5+11/2*2,5+7/10`
`=0,8+2,2+0,7`
`=3+0,7=3,7`
`b)125%xx 17/4:(1 5/16-0,5)+2008`
`=1,25xx4,25:13/16+2008`
`=85/13+2008`
`=2014 7/13`
`c)5/11+(16/11+1)`
`=5/11+1+5/11+1`
`=2+10/11=32/11`
`d)3/17+11/4+5/8+14/17+3/8`
`=3/17+14/17+5/8+3/8+11/4`
`=1+1+11/4`
`=19/4`
a)
\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)
<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)
<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)
b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)
c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)
= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)
d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)
= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a/`
` 7/5 + 3 2/5 - 1 1/2 `
`= (7/5 + 17/5) - 3/2`
`= 24/5 - 3/2 `
`= 48/10 - 15/10 `
`= 33/10 `
`b/`
` 3 xx 2 4/9 xx 3/2 `
` = 3 xx 22/9 xx 3/2 `
` = 22/3 xx 3/2`
`= 11.`
`c/`
` 5/9 xx ( 2 1/6 - 1 2/3 ) `
`= 5/9 xx ( 13/6 -5/3 )`
`= 5/9 xx ( 13/6 - 10/6 ) `
`= 5/9 xx 3/6 `
`= 5/9 xx 1/2 `
`= 5/18`
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$
$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$
$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$
Bài 2:
$18: \frac{x\times 0,4+0,32}{x}+5=14$
$18: \frac{x\times 0,4+0,32}{x}=14-5=9$
$\frac{x\times 0,4+0,32}{x}=18:9=2$
$x\times 0,4+0,32=2\times x$
$2\times x-x\times 0,4=0,32$
$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$
![](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)
Tham khảo link: https://olm.vn/hoi-dap/detail/55111422944.html
`(x+1/3)+(x+1/9)+(x+1/27)+(x+1/81)=56/81`
`x+x+x+x+1/3+1/9+1/27=56/81-1/81`
`4x+13/27=55/81`
`4x=55/81-13/27`
`4x=55/81-52/81`
`4x=16/81`
`x=4/108`
Vậy `x=4/108`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(1-\dfrac{1}{x+3}=\dfrac{34}{103}:\dfrac{1}{3}=\dfrac{34}{103}.3\)
\(1-\dfrac{1}{x+3}=\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=1-\dfrac{102}{103}=\dfrac{103}{103}-\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=\dfrac{1}{103}\)
\(\Rightarrow x+3=103\)
\(x=103-3\)
\(x=100\)
Vậy x = 100
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{\left(\dfrac{5}{30}+\dfrac{3}{30}+\dfrac{2}{30}\right):\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{2}{30}\right)}{\left(\dfrac{30}{60}-\dfrac{20}{60}+\dfrac{15}{60}-\dfrac{12}{60}\right):\left(\dfrac{3}{12}-\dfrac{2}{12}\right)}=\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{1}{3}\times5}{\dfrac{13}{60}\times12}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)
Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
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