1005/1008=?
K
Khách
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MN
1
26 tháng 4 2018
Đặt \(A=\frac{1005}{1006}+\frac{1006}{1007}+\frac{1007}{1008}+\frac{1008}{1005}\) ta có :
\(A=\frac{1006-1}{1006}+\frac{1007-1}{1007}+\frac{1008-1}{1008}+\frac{1005+3}{1005}\)
\(A=\frac{1006}{1006}-\frac{1}{1006}+\frac{1007}{1007}-\frac{1}{1007}+\frac{1008}{1008}-\frac{1}{1008}+\frac{1005}{1005}+\frac{3}{1005}\)
\(A=1-\frac{1}{1006}+1-\frac{1}{1007}+1-\frac{1}{1008}+1+\frac{3}{1005}\)
\(A=\left(1+1+1+1\right)-\left(\frac{1}{1006}+\frac{1}{1007}+\frac{1}{1008}-\frac{3}{1005}\right)\)
\(A=4-\left(\frac{1}{1006}+\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1005}-\frac{1}{1005}-\frac{1}{1005}\right)\)
\(A=4-\left[\left(\frac{1}{1006}-\frac{1}{1005}\right)+\left(\frac{1}{1007}-\frac{1}{1005}\right)+\left(\frac{1}{1008}-\frac{1}{1005}\right)\right]\)
Mà :
\(\frac{1}{1006}< \frac{1}{1005}\)\(\Rightarrow\)\(\frac{1}{1006}-\frac{1}{1005}< 0\) \(\left(1\right)\)
\(\frac{1}{1007}< \frac{1}{1005}\)\(\Rightarrow\)\(\frac{1}{1007}-\frac{1}{1005}< 0\) \(\left(2\right)\)
\(\frac{1}{1008}< \frac{1}{1005}\)\(\Rightarrow\)\(\frac{1}{1008}-\frac{1}{1005}< 0\) \(\left(3\right)\)
Từ (1), (2) và (3) suy ra :
\(\left(\frac{1}{1006}-\frac{1}{1005}\right)+\left(\frac{1}{1007}-\frac{1}{1005}\right)+\left(\frac{1}{1008}-\frac{1}{1005}\right)< 0\)
\(\Rightarrow\)\(A=4-\left[\left(\frac{1}{1006}-\frac{1}{1005}\right)+\left(\frac{1}{1007}-\frac{1}{1005}\right)+\left(\frac{1}{1008}-\frac{1}{1005}\right)\right]>4\)
\(\Rightarrow\)\(A>4\) ( điều phải chứng minh )
Vậy \(A>4\)
Chúc bạn học tốt ~
31 tháng 8 2021
\(\dfrac{1004}{1005}< \dfrac{1005}{1006}< \dfrac{1006}{1007}< \dfrac{1007}{1008}\)
M
1
26 tháng 1 2021
Ta có: \(\dfrac{x+1006}{1007}+\dfrac{x+1005}{1008}=\dfrac{x+1004}{1009}+\dfrac{x+1003}{1010}\)
\(\Leftrightarrow\dfrac{x+1006}{1007}+1+\dfrac{x+1005}{1008}+1=\dfrac{x+1004}{1009}+1+\dfrac{x+1003}{1010}+1\)
\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}=\dfrac{x+2013}{1009}+\dfrac{x+2013}{1010}\)
\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}-\dfrac{x+2013}{1009}-\dfrac{x+2013}{1010}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\right)=0\)
mà \(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\ne0\)
nên x+2013=0
hay x=-2013
Vậy: S={-2013}
LX
2
KN
3 tháng 3 2020
\(\frac{x+6}{1005}+\frac{x+132}{471}+\frac{x+1008}{168}=-12\)
\(\Leftrightarrow\frac{x+6}{1005}+2+\frac{x+132}{471}+4+\frac{x+1008}{168}+6=0\)
\(\Leftrightarrow\frac{x+2016}{1005}+\frac{x+2016}{471}+\frac{x+2016}{168}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{1005}+\frac{1}{471}+\frac{1}{168}\right)=0\)
Dễ thấy \(\frac{1}{1005}+\frac{1}{471}+\frac{1}{168}>0\)nên x + 2016 = 0
Vậy x = -2016
3 tháng 3 2020
\(\frac{x+6}{1005}+\frac{x+132}{471}+\frac{x+1008}{168}=-12\)
\(\Leftrightarrow\frac{x+6}{3\cdot335}+\frac{x+132}{3\cdot157}+\frac{x+1008}{3\cdot56}=-12\)
\(\Leftrightarrow\frac{x+6}{335}+\frac{x+132}{157}+\frac{x+1008}{56}=-36\)
\(\Leftrightarrow\frac{x}{335}+\frac{x}{157}+\frac{x}{56}+\frac{6}{335}+\frac{132}{157}+18=-36\)
\(\Leftrightarrow\frac{x}{335}+\frac{x}{157}+\frac{x}{56}=-54-\frac{6}{335}-\frac{132}{157}\)
\(\Leftrightarrow x\left(\frac{1}{335}+\frac{1}{157}+\frac{1}{56}\right)=-6-\frac{6}{335}-12-\frac{132}{157}-36\)
\(\Leftrightarrow x\left(\frac{1}{335}+\frac{1}{157}+\frac{1}{56}\right)=\frac{-2016}{335}+\frac{-2016}{157}+\frac{-2016}{56}\)
\(\Leftrightarrow x\left(\frac{1}{335}+\frac{1}{157}+\frac{1}{56}\right)=-2016\left(\frac{1}{335}+\frac{1}{157}+\frac{1}{56}\right)\)
\(\Leftrightarrow x=-2016\)
3 tháng 3 2017
\(\frac{x+1006}{1007}+\frac{x+1005}{1008}=\frac{x+1004}{1009}+\frac{x+1003}{1010}\)
\(\Rightarrow\left(\frac{x+1006}{1007}+1\right)+\left(\frac{x+1005}{1008}+1\right)=\left(\frac{x+1004}{1009}+1\right)+\left(\frac{x+1003}{1010}+1\right)\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}=\frac{x+2013}{1009}+\frac{x+2013}{1010}\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}-\frac{x+2013}{1009}-\frac{x+2013}{1010}=0\)
\(\Rightarrow\left(x+2013\right)\left(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\right)=0\)
Mà \(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\ne0\)
\(\Rightarrow x+2013=0\)
\(\Rightarrow x=-2013\)
Vậy x = -2013
0.9970238095
\(\frac{335}{336}\)
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