a) 1/3 + 1/6 + 1/12 + 1/24 + 1/48 + 1/96 + 1/192
b) 2,3 x 1,3 – 2,3 x 1,2 – 2,3 x 0,09
K
Khách
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2 tháng 8 2019
\(x\cdot\frac{1}{1\cdot2}+x\cdot\frac{1}{2\cdot3}+x\cdot\frac{1}{3\cdot4}+...+x\cdot\frac{1}{9\cdot10}=2\)
\(\Leftrightarrow x\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)=2\)
\(\Leftrightarrow x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)=2\)
\(\Leftrightarrow x\cdot\left(1-\frac{1}{10}\right)=2\)
\(\Leftrightarrow x\cdot\frac{9}{10}=2\)
\(\Leftrightarrow x=2\cdot\frac{10}{9}=\frac{20}{9}\)
HP
6
NK
4
HT
3
WJ
23 tháng 10 2015
1)Ta có: \(12,\left(1\right)=12+0,\left(1\right)=12+\frac{1}{9}=\frac{109}{9}\);
\(2,3\left(6\right)=2,3+\frac{1}{10}\times0,\left(6\right)=2,3+\frac{1}{10}\times6\times0,\left(1\right)=2,3+\frac{1}{10}\times6\times\frac{1}{9}=\frac{71}{30}\)\(4,\left(21\right)=4+21\times0,\left(01\right)=4+21\times\frac{1}{99}=\frac{139}{33}\)
\(\Rightarrow\)\(\left[\frac{109}{9}-\frac{71}{30}\right]\div\frac{139}{33}=\frac{9647}{4170}\)
2)Ta có: \(0,\left(12\right)=12\times0,\left(01\right)=12\times\frac{1}{99}=\frac{4}{33}\)
\(1,\left(6\right)=1+6\times0,\left(1\right)=1+6\times\frac{1}{9}=\frac{5}{3}\)
\(0,\left(4\right)=4\times0,\left(1\right)=4\times\frac{1}{9}=\frac{4}{9}\)
\(\Rightarrow\frac{4}{33}\div\frac{5}{3}=x\div\frac{4}{9}\Rightarrow x\div\frac{4}{9}=\frac{4}{55}\Rightarrow x=\frac{4}{55}\times\frac{4}{9}\Rightarrow x=\frac{16}{495}\)
28 tháng 3 2017
1/3 + 1/6 + 1/12 + 1/24 + 1/48 + 1/96 + 1/192
= 127/192
k minh di , xin day
15 tháng 8 2021
Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Leftrightarrow100\cdot\dfrac{9}{10}-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Leftrightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=1\)
\(\Leftrightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=2\)
\(\Leftrightarrow x=-\dfrac{81}{100}\)
NM
1
AH
Akai Haruma
Giáo viên
30 tháng 4 2023
Lời giải:
$1,2+2,3+3,4+...+98,99+99,100=2,4x-1$
$(1+0,2)+(2+0,3)+(3+0,4)+...+(98+0,99)+(99+0,100)=2,4x-1$
$(1+2+3+...+98+99)+(0,2+0,3+0,4+...+0,99+0,100)=2,4x-1$
$=99.100:2 + (0,2+0,3+...+0,9)+(0,10+0,11+0,12+...+0,99)+0,1=2,4x-1$
$4950+0,1(2+3+...+9)+0,01(10+11+...+99)+0,1=2,4x-1$
$4950+0,1.44+0,01.4905+0,1=2,4x-1$
$5003,55=2,4x-1$
$x=\frac{100091}{48}$
a) 1/3 + 1/6 + 1/12 + 1/24 + 1/48 + 1/96 + 1/192
= 64/192 + 32/192 + 8/192 + 4/192 + 2/192 + 1/192
= 64 + 32 + 8 + 4 + 2 + 1 / 192
= 111/192
= 37/64
b) 2,3 x 1,3 – 2,3 x 1,2 – 2,3 x 0,09
= 2,3 x ( 1,3 - 1,2 -0,09)
= 2,3 x 0,01
= 0,023