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....

\(-x=\frac{357}{352}=1+\frac{5}{352}\)

\(-y=\frac{1000}{999}=1+\frac{1}{999}\)

\(\frac{5}{352}>\frac{5}{999}>\frac{1}{999}\)

\(=>\frac{357}{352}>\frac{1000}{999}=>-x>-y\)

\(=>x< y\)

biểu thức 2 lớn hơn ạ!

làm như nào b ơi

Cộng cả x và y với 1 ta được

x + 1 = \(\frac{-357}{352}+1=\frac{-5}{352}\)< \(\frac{-1}{352}\)

y + 1 = \(\frac{-1000}{999}+1=\frac{-1}{999}\)>\(\frac{-1}{352}\)

Như vậy x + 1 < y + 1 hay x < y

Ta có: \(-\frac{1}{5}< 0\)

\(\frac{1}{1000}>0\)

nên \(-\frac{1}{5}< \frac{1}{1000}\)

\(\frac{-1}{5}\)<0 vì có tử hoặc mẫu âm

\(\frac{1}{1000}\)>0 vì k có tử hoặc mẫu âm

vậy \(\frac{-1}{5}\) < \(\frac{1}{1000}\)

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 \(C< \frac{1}{3}\)

\(\text{Ta thấy: }18>17\Rightarrow\frac{18}{17}>1\Rightarrow\frac{-18}{17}<-1\)

\(999<1000\Rightarrow\frac{999}{1000}<1\Rightarrow\frac{-999}{1000}>-1\)

\(\text{Vậy }\frac{-18}{17}<\frac{-999}{1000}\)

a) Ta có:

-1/10 < 0

1/1000 > 0

=> -1/10 < 1/1000

b) Ta có:

357/358 < 1

1000 / 999  > 1

=> 357/358  < 1000/999

=> -357/358  > -1000/999

c) -151515/313131 = -15/31

Vậy -15/13 = -151515/313131