Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có \(\frac{{ - 2}}{3} < 0\) và \(\frac{1}{{200}} > 0\) nên \(\frac{{ - 2}}{3}\)<\(\frac{1}{{200}}\).
b) Ta có: \(\frac{{139}}{{138}} > 1\) và \(\frac{{1375}}{{1376}} < 1\) nên \(\frac{{139}}{{138}}\) > \(\frac{{1375}}{{1376}}\).
c) Ta có: \(\frac{{ - 11}}{{33}} = \frac{{ - 1}}{3}\) và \(\frac{{25}}{{ - 76}} = \frac{{ - 25}}{{76}} > \frac{{ - 25}}{{75}} = \frac{{ - 1}}{3}\,\,\,\, \Rightarrow \frac{{25}}{{ - 76}} > \frac{{ - 11}}{33}\).
a: -2/3<0<1/200
b: 139/138>1
1375/1376<1
=>139/138>1375/1376
c: -11/33=-1/3=-25/75<-25/76
a) Ta có: \(\frac{2}{{ - 5}} = \frac{{ - 16}}{{40}}\) và \(\frac{{ - 3}}{8} = \frac{{ - 15}}{{40}}\)
Do \(\frac{{ - 16}}{{40}} < \frac{{ - 15}}{{40}}\,\, \Rightarrow \,\frac{2}{{ - 5}} < \frac{{ - 3}}{8}\).
b) Ta có: \( - 0,85 = \frac{{ - 85}}{{100}} = \frac{{ - 17}}{{20}}\). Vậy \( - 0,85\)=\(\frac{{ - 17}}{{20}}\).
c) Ta có: \(\frac{{37}}{{ - 25}} = \frac{{ - 296}}{{200}}\)
Do \(\frac{{ - 137}}{{200}} > \frac{{ - 296}}{{200}}\) nên \(\frac{{ - 137}}{{200}}\) > \(\frac{{37}}{{ - 25}}\) .
d) Ta có: \( - 1\frac{3}{{10}}=\frac{-13}{10}\) ;
\(-\left( {\frac{{ - 13}}{{ - 10}}} \right) = \frac{{-13}}{{10}}\).
Vậy \(- 1\frac{3}{{10}} =-(\frac{{-13}}{{-10}})\,\).
\(B=\frac{1}{199}+\frac{2}{198}+...+\frac{199}{1}\)
\(=1+\frac{1}{199}+1+\frac{2}{198}+...+\frac{199}{1}+1-199\)
\(=200+\frac{200}{2}+...+\frac{200}{199}-199\)
\(=1+\frac{200}{2}+...+\frac{200}{199}\)
\(=200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{200}\)
Ta có:
A=-2012/4025=>-2012/4025x2=-4024/4025
B=-1999/3997=>-1999/3997x2=-3998/3997
Ta có: 4024/4025<1<3998/3997
=>4024/4025<3998/3997
=>-4024/4025>-3998/3997
=>-2012/4025>-1999/3997
a) vì -2005/2006 <0<1/200 nên -2005/2006<1/200
b)vì 1/4003>0>-75/106 nên 1/4003>-75/106
câu cuôi mình đang bí co gì mình giuuwi câu trả lời sau
a) vì -2005/2006<-1<1/200 nên -2005/2006<1/200
b) vì 1/4003>-1>-75/106 nên 1/4003>-75/106
c) vì 1250/1251<1<25/24 nên 1250/1251<25/24
Ta có -2012/4025 < -2012/4024 tức là < -1/2
Ta có -1999/3997 > -1999/3998 tức là > -1/2
=> -1999/3997 > -2012/4025
Ta có 3^21 = 3^(2.10 + 1) = 9^ 10 .3
Ta có 2^31= 2^( 3.10+1) = 8^10.2
Từ đó => 3^21 > 2^31
a) Ta có: \(\frac{2012}{4025}< \frac{2012}{4024}=\frac{1}{2}\)
mà \(\frac{1999}{3997}>\frac{1999}{3998}=\frac{1}{2}\)
\(\Rightarrow\frac{2012}{4025}< \frac{1999}{3997}\)\(\Rightarrow\frac{-2012}{4025}>\frac{-1999}{3997}\)\(\Rightarrow A>B\)
b) \(A=3^{21}=3^{20+1}=3^{20}.3=3^{2.10}.3=9^{10}.3\)
\(B=2^{31}=2^{30+1}=2^{30}.2=2^{3.10}.2=8^{10}.2\)
Vì \(9>8\)\(\Rightarrow9^{10}>8^{10}\)
mà \(3>2\)\(\Rightarrow9^{10}.3>8^{10}.2\)\(\Rightarrow3^{21}>2^{31}\)\(\Rightarrow A>B\)
a) ta có: \(1+\frac{-21}{25}=\frac{4}{25}=\frac{44}{375};1+\frac{-123}{167}=\frac{44}{167}.\)
\(\Rightarrow\frac{44}{375}< \frac{44}{167}\Rightarrow1+\frac{-21}{25}< 1+\frac{-123}{375}\Rightarrow\frac{-21}{25}< \frac{-123}{375}\)
b) ta có: \(\frac{199}{200}=\frac{1990}{2000}\)
=> ...
\(\frac{-123}{167}\)lớn hơn \(\frac{-21}{25}\)
\(\frac{199}{200}\)nhỏ hơn\(\frac{1999}{2000}\)