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Ta có: 20082 > 20082 -1 = 2007.2009 => 1007/1008<1008/1009
Ta có:\(\frac{1007}{1008}=1-\frac{1}{1008}\)
\(\frac{1008}{1009}=1-\frac{1}{1009}\)
mà \(\frac{1}{1008}>\frac{1}{1009}\)
=> \(1-\frac{1}{1008}< 1-\frac{1}{1009}\)
Hay \(\frac{1007}{1008}< \frac{1008}{1009}\)
Vậy .......
Chúc bạn hk tốt!!! nhớ k cho mình na
\(\dfrac{4}{17}=\dfrac{16}{68}\\ Vì:\dfrac{16}{68}< \dfrac{16}{63}\Rightarrow\dfrac{4}{17}< \dfrac{16}{63}\\ ---\\ \dfrac{1007}{1009}=1-\dfrac{2}{1009};\dfrac{1005}{1007}=1-\dfrac{2}{1007}\\ Vì:\dfrac{2}{1009}< \dfrac{2}{1007}\Rightarrow1-\dfrac{2}{1009}>1-\dfrac{2}{1007}\\ \Rightarrow\dfrac{1007}{1009}>\dfrac{1005}{1007}\)
a: 4/17=16/68
16/68<16/63
=>4/17<16/63
b: 19/53<20/53
20/53<20/50(Vì 53>50)
=>19/53<20/50=2/5
mà 2/5=30/75<30/73
nên 19/53<30/73
c: 1007/1009=1-2/1009
1005/1007=1-2/1007
1009>1007
=>2/1009<2/1007
=>-2/1009>-2/1007
=>1007/1009>1005/1007
\(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+.....+\dfrac{x+1007}{1008}=\dfrac{x+1008}{1007}+\dfrac{x+1009}{1006}+........+\dfrac{x+2014}{1}\)\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+...+\left(\dfrac{x+1007}{1008}+1\right)=\left(\dfrac{x+1008}{1007}+1\right)+\left(\dfrac{x+1009}{1006}+1\right)+...+\left(\dfrac{x+2014}{1}+1\right)\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+1007}{1008}=\dfrac{x+2015}{1007}+\dfrac{x+1009}{1006}+...+\dfrac{x+2014}{1}\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+2015}{1008}-\dfrac{x+1008}{1007}-\dfrac{x+2015}{1006}-...-\dfrac{x+2015}{1}=0\)\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1\right)=0\)\(\Leftrightarrow x+2015=0\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1>0\right)\)\(\Leftrightarrow x=-2015\)
Vậy x=-2015
Ta có:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
=\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
=\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1006}\right)\)
=\(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}\)
=>\(\left(\frac{A}{B}\right)^{2013}\)=(\(\frac{\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}}{\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}}^{ }\))2013=12013=1
\(=\frac{2016.2017+2018.21+1996}{2017.2016+2018.2017}\)
\(=\frac{2018.21+1996}{2018.2017}\)
\(=\frac{21+1996}{2017}\)
\(=\frac{2017}{2017}\)
\(=1\)
C = 1.2.3.4....2011
D = 1007/2 . 1008/2 . 1009/2.....2012/2
D = (1007.1008.1009.....2012) : (2.2.2.2........2) (có 2012 - 1007 + 1 = 1006 số 2 )
D = (1007.1008.1009....503 .2.2) : 21004
MÀ (4.503.1007.1008.....2011) < (1.2.3.....2011)
Vậy c > D
Trả lời
A=4 067 281 - 4 067 281
=0
Học tốt !
A= (2017.2016+1009)-(2018.2016-1007)
A=2016x(2017-2018)+1009+1007
A=2016x-1+(1009+1007)
A=-2016+2016
A=0