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1\(\dfrac{1}{12}x1\dfrac{1}{13}x1\dfrac{1}{14}\) x ... x \(1\dfrac{1}{2005}\)
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TC
26 tháng 3 2022
\(=>C=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}.....\cdot\dfrac{101}{100}\)
\(C=\dfrac{3\cdot4\cdot5.......\cdot101}{2\cdot3\cdot4.........\cdot100}\)
\(C=\dfrac{101}{2}\)
ND
Nguyễn Đức Trí
VIP
3 tháng 8 2023
\(...=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}....x\dfrac{1000}{999}\)
\(=\dfrac{1}{2}x\dfrac{1000}{1}=500\)
3 tháng 8 2023
=3/2x4/3x5/4x....x1000/999
=1/2x1000=500
mình chưa chắc là đúng đâu nhé
O
2 tháng 4 2022
a) \(\dfrac{-5}{9}-\dfrac{-5}{12}=\dfrac{-5}{9}+\dfrac{5}{12}=\dfrac{-20}{36}+\dfrac{15}{36}=-\dfrac{5}{36}\)
b) \(\dfrac{-5}{12}:\dfrac{15}{4}=\dfrac{-5}{12}\times\dfrac{4}{15}=\dfrac{-1}{9}\)
c) \(\dfrac{1}{13}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{13}-\dfrac{14}{13}=\dfrac{1}{13}\cdot\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{14}{13}=\dfrac{1}{13}\cdot1-\dfrac{14}{13}=\dfrac{1}{13}-\dfrac{14}{13}=-1\)
AD
4
5 tháng 7 2023
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)\(\left(x+1\right)\times\dfrac{1}{10}+\left(x+1\right)\times\dfrac{1}{11}+\left(x+1\right)\times\dfrac{1}{12}-\left(x+1\right)\times\dfrac{1}{13}-\left(x+1\right)\times\dfrac{1}{14}=0\)
\(\left(x+1\right)\times\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}>0\)
=> \(x+1=0\)
\(x=0-1\)
\(x=-1\)
DD
5 tháng 7 2023
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\\ \Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\\ \Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\\ \Rightarrow x+1=0\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\\ \Rightarrow x=-1\)
NL
1
AH
Akai Haruma
Giáo viên
16 tháng 9 2023
Lời giải:
$\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$
$\Rightarrow (x+1)(\frac{1}{10}+\frac{1}{11}+\frac{1}{12})=(x+1)(\frac{1}{13}+\frac{1}{14})$
$\Rightarrow (x+1)(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14})=0$
Hiển nhiên $\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0$
$\Rightarrow x+1=0$
$\Rightarrow x=-1$
13 tháng 5 2023
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 5 2023
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
8 tháng 6 2021
Ta có: A\(=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}=\dfrac{2}{45}\)
8 tháng 6 2021
\(A=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}\)
\(=\dfrac{2}{45}\)
-Chúc bạn học tốt-
25 tháng 3 2022
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)
Vậy \(x\in\left\{22;23;24;...\right\}\)
L
3
QN
20 tháng 8 2017
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}=\dfrac{x+1}{14}+\dfrac{x+1}{15}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}\right)=\left(x+1\right)\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
20 tháng 8 2017
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}=\dfrac{x+1}{14}+\dfrac{x+1}{15}\)
<=> \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}-\dfrac{x+1}{14}-\dfrac{x+1}{15}=0\)
<=> \(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
Do: \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{14}>0\) nên x + 1 = 0
Vậy x = -1
1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\)...\(\times\)1\(\dfrac{1}{2005}\)
A = \(\dfrac{12+1}{12}\) \(\times\) \(\dfrac{13+1}{13}\) \(\times\) \(\dfrac{14+1}{14}\)\(\times\)...\(\times\) \(\dfrac{2006}{2005}\)
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\)...\(\times\) \(\dfrac{2006}{2005}\)
A = \(\dfrac{2006}{12}\)
A = \(\dfrac{1003}{6}\)