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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 ...
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....
\(\text{Ta thấy: }18>17\Rightarrow\frac{18}{17}>1\Rightarrow\frac{-18}{17}<-1\)
\(999<1000\Rightarrow\frac{999}{1000}<1\Rightarrow\frac{-999}{1000}>-1\)
\(\text{Vậy }\frac{-18}{17}<\frac{-999}{1000}\)
a ) \(-5=\frac{-5}{1}< 0\)và \(\frac{1}{63}>0\)
\(\Rightarrow-5< \frac{1}{63}\)
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
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 \(C< \frac{1}{3}\)