Tìm x biết: \(\dfrac{1}{2}\) : (1+\(\dfrac{1}{2}\)) : (1+\(\dfrac{1}{3}\)) : ... : (1+\(\dfrac{1}{x}\)) = \(\dfrac{1}{101}\)
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HH
1
14 tháng 2 2022
x + 1/100 + x + 2/101 = x + 3/102 - 1
<=> x + 1/100 - 1 + x + 2/101 - 1 = x + 3/102 - 1 - 2
<=> x - 99/100 + x - 99/101 = x - 99/102 - 2
<=> x - 99/100 + x - 99/101 - x - 99/102 = -2
<=> (x - 99)(1/100 + 1/101 - 1/102) = -2
<=> x - 99 = -2/1/100 + 1/101 - 1/102
<=> x = -2/1/100 + 1/101 - 1/102 + 99
Bạn chịu khó bấm máy hộ mình, số to quá
17 tháng 3 2018
a/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^{10}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^9}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{10}}\)
b/ \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+.......+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Leftrightarrow3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+......+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{101}{1540}.3\)
\(\Leftrightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+......+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+.....+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Leftrightarrow x+3=308\)
\(\Leftrightarrow x=305\)
Vậy ..
c/ \(1+\dfrac{1}{3}+\dfrac{1}{6}+........+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{2007}{2009}\)
\(\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{6}+.......+\dfrac{1}{x\left(x+1\right):2}\right)=\dfrac{4016}{2009}.\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+......+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)
\(\Leftrightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+......+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2009}\)
\(\Leftrightarrow x+1=2009\)
\(\Leftrightarrow x=2008\)
Vậy ..
17 tháng 3 2018
bài 1:
A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
ta thấy 2A=\(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^9}\)
=>2A-A=\(1-\dfrac{1}{2^{10}}=\dfrac{1023}{1024}\)
4 tháng 2 2021
a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)
\(=\dfrac{1}{2016}\)
b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)
\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)
hay x=10
Vậy: x=10
4 tháng 5 2021
Giải:
a)1/5.8+1/8.11+...+1/x.(x+1)=101/1540
1/3.(3/5.8+3/8.11+...+3/x.(x+1))=101/1540
1/3.(1/5-1/8+1/8-1/11+...+1/x-1/x+1)=101/1540
1/3.(1/5-1/x+1)=101/1540
1/5-1/x+1=101/1540:1/3
1/5-1/x+1=303/1540
1/x+1=1/5-303/1540
1/x+1=1/308
⇒x+1=308
x=308-1
x=307
b)1/1.2+1/2.3+1/3.4+...+1/x.(x+1)=2020/2021
1/1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2020/2021
1/1-1/x+1=2020/2021
1/x+1=1/1-2020/2021
1/x+1=1/2021
⇒x+1=2021
x=2021-1
x=2020
Mk thấy đề bài hơi sai là:
1/x+(x+1) ➜ 1/x.(x+1)
mới ra đc kết quả!
4 tháng 5 2021
cảm ơn bn đã cố gắng
à bn đã tham gia khóa học của mình chưa
12 tháng 3 2022
a)4/5+x=2/3
x=2/3-4/5
x=-2/15
b)-5/6-x=2/3
x=-5/6-2/3
x=-3/2
c)1/2x+3/4=-3/10
1/2x=-3/10-3/4
1/2x=-21/20
x=-21/20:1/2
x=-21/10
d)x/3-1/2=1/5
x/3=1/5+1/2
x/3=7/10
10x/30=21/30
10x=21
x=21:10
x=21/10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 9 2023
a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)
- \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) < \(x\) < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)
- \(\dfrac{46}{3}\) < \(x\) < - \(\dfrac{13}{7}\)
\(x\) \(\in\) {-15; -14;-13;..; -2}
23 tháng 9 2023
a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)
Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)
Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)
Suy ra \(-15\le x\le-2\), x ϵ Z
b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)
Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)
Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)
Suy ra \(-1\le x\le0\), x ϵ Z
HK
16 tháng 4 2023
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))
vậy x= 2023
NH
13 tháng 4 2023
sửa đề: phải là 14 chứ sao lại là 13 nhỉ?=))
\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\) \(\left(x\ne0;x\ne-3\right)\)
\(\left(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{x\left(x+3\right)}\right)\cdot3=\dfrac{101}{1540}\cdot3\)
\(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{308\left(x+3\right)}{1540\left(x+3\right)}-\dfrac{1540}{1540\left(x+3\right)}=\dfrac{303\left(x+3\right)}{1540\left(x+3\right)}\)
suyy ra
`308x+924-1540=303x+909`
`5x=1525`
`x=305(tm)`
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 9 2023
a, - \(\dfrac{1}{10}\) + \(\dfrac{2}{5}\)\(x\) + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\)\(x\) = \(\dfrac{1}{10}\) - \(\dfrac{7}{20}\) + \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\) \(x\) = - \(\dfrac{3}{20}\)
\(x\) = - \(\dfrac{3}{20}\): \(\dfrac{2}{5}\)
\(x\) = - \(\dfrac{3}{8}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 9 2023
b, \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\) - \(\dfrac{1}{3}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{8}{15}\)
\(x\) = \(\dfrac{1}{2}\): (- \(\dfrac{8}{15}\))
\(x\) = - \(\dfrac{15}{16}\)
`1/2:(1+1/2):(1+1/3):...:(1+1/x)=1/101`
`<=>1/2:3/2:4/3:...:[x+1]/x=1/101`
`<=>1/2 . 2/3 . 3/4 . .... . x/[x+1]=1/101`
`<=>1/[x+1]=1/101`
`<=>x+1=101`
`<=>x=100`