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26 tháng 2 2019
\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+\frac{2016}{998}+...+\frac{2016}{501}}{-\frac{1}{1\cdot2}-\frac{1}{3\cdot4}-\frac{1}{5\cdot6}-...-\frac{1}{999\cdot1000}}\)
\(B=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{999\cdot1000}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{500}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}\)
\(B=\frac{2016}{-1}=-2016\)
LH
1
13 tháng 12 2017
Ta so sánh hai phân số \(\frac{2010}{2011}\)và \(\frac{1000}{999}\)có :
\(\frac{2010}{2011}< \frac{1000}{999}\)
\(\Rightarrow\left(\frac{-2010}{2011}\right)>\left(\frac{-1000}{999}\right)\)
Vậy ...
TM
2
25 tháng 8 2018
Ta có \(x=\frac{357}{-352}\)
\(\Rightarrow-x=\frac{357}{352}=1+\frac{2}{352}=\frac{1}{176}\)
Ta có \(y=\frac{-1000}{999}\)
\(\Rightarrow-y=\frac{1000}{999}=1+\frac{1}{999}\)
Vì \(\frac{1}{176}>\frac{1}{999}\Rightarrow1+\frac{1}{176}>1+\frac{1}{999}\Rightarrow-x>-y\Rightarrow x< y\)
Khi đó x < y
Vậy....
TL
1
28 tháng 8 2015
\(\text{Ta thấy: }18>17\Rightarrow\frac{18}{17}>1\Rightarrow\frac{-18}{17}<-1\)
\(999<1000\Rightarrow\frac{999}{1000}<1\Rightarrow\frac{-999}{1000}>-1\)
\(\text{Vậy }\frac{-18}{17}<\frac{-999}{1000}\)
TM
1
IM
27 tháng 11 2016
Ta có :
\(C=\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\)
\(\Rightarrow4C=1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\)
\(\Rightarrow4C-C=\left(1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\right)\)
\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)
\(\Rightarrow C=\frac{1}{3}-\frac{1}{3.4^{1000}}< \frac{1}{3}\)
=> C < 1 / 3
27 tháng 11 2016
Ta có:
\(C=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\)
\(\Rightarrow4C=1+\frac{1}{4}+...+\frac{1}{4^{999}}\)
\(\Rightarrow4C-C=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}\right)\)
\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)
\(\Rightarrow C=\left(1-\frac{1}{4^{1000}}\right).\frac{1}{3}\)
\(\Rightarrow C=\frac{1}{3}-\frac{1}{4^{1000}.3}\)
Mà \(\frac{1}{3}>\frac{1}{3}-\frac{1}{4^{1000}.3}\)
\(\Rightarrow C< \frac{1}{3}\)
Vậy \(C< \frac{1}{3}\)