SO SANHS CÁC SỐ HỮU TỈ SAU:
a/\(\frac{325}{260}\)và\(\frac{876}{320}\)
b/\(\frac{13}{9}\)và\(\frac{8}{3}\)
c/\(\frac{18}{-39}\)và\(\frac{-17}{41}\)
d/\(\frac{-15}{25}\)và\(\frac{-151515}{232323}\)
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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 \(- 1\frac{3}{{10}} =-(\frac{{-13}}{{-10}})\,\).
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: \(\dfrac{325}{260}< 2< \dfrac{876}{320}\) => \(\dfrac{325}{260}< \dfrac{876}{320}\)
b) Ta có: \(\dfrac{13}{9}< \dfrac{24}{9}=\dfrac{8}{3}\) => \(\dfrac{13}{9}< \dfrac{8}{3}\)
c) Ta có: \(\dfrac{18}{-39}=\dfrac{-18}{39}< \dfrac{-17}{39}< \dfrac{-17}{41}\) => \(\dfrac{18}{-39}< \dfrac{-17}{41}\)
d) Ta có: \(\dfrac{-151515}{232323}=\dfrac{-151515\div10101}{232323\div10101}=\dfrac{-15}{23}\)
Mà \(\dfrac{-15}{25}>\dfrac{-15}{23}\) => \(\dfrac{-15}{25}>\dfrac{-151515}{232323}\)
a) \(\dfrac{325}{260}=\dfrac{5}{4}=\dfrac{100}{80}< \dfrac{219}{80}=\dfrac{876}{320}\)
b) \(\dfrac{13}{9}< \dfrac{24}{9}=\dfrac{8}{3}\)
c) \(\dfrac{18}{-39}=\dfrac{-738}{1599}< \dfrac{-663}{1599}=\dfrac{-17}{41}\)
d) \(\dfrac{-15}{25}=\dfrac{-151515}{252525}< \dfrac{-151515}{232323}\)
a.\(\frac{13}{17}\)=1-\(\frac{4}{17}\); \(\frac{46}{50}\)=1-\(\frac{4}{50}\)
Vì \(\frac{4}{17}\)>\(\frac{4}{50}\)=> 1-\(\frac{4}{17}\)<1-\(\frac{4}{50}\)
Vậy\(\frac{13}{17}\)<\(\frac{46}{50}\)
a)
\(\frac{13}{27}=\frac{13.101}{27.101}=\frac{1313}{2727}\)
=> \(\frac{13}{27}=\frac{1313}{2727}\)
b)
\(-\frac{15}{23}=-\frac{15.10101}{23.10101}=-\frac{151515}{232323}\)
=>\(-\frac{15}{23}=-\frac{151515}{232323}\)
a) \(\frac{1313}{2727}=\frac{1313:101}{2727:101}=\frac{13}{27}\)
Vậy \(\frac{13}{27}=\frac{1313}{2727}\)
b) \(-\frac{151515}{232323}=\frac{-151515:10101}{232323:10101}=-\frac{15}{23}\)
Vậy \(-\frac{15}{23}=-\frac{151515}{232323}\)
a/ >
b/ <
c/ <
d/ =
a) \(\frac{325}{260}=\frac{650}{520}< \frac{650}{320}< \frac{876}{320}\)
\(\Rightarrow\frac{325}{260}< \frac{876}{320}.\)
b) \(\frac{13}{9}< \frac{24}{9}=\frac{8}{3}\)
\(\Rightarrow\frac{13}{9}< \frac{8}{3}.\)
c) Do \(18>-17;-39< 41\)
\(\Rightarrow\frac{18}{-39}>\frac{-17}{41}.\)
d) \(\frac{-151515}{232323}=\frac{\left(-15\right).10101}{23.10101}=\frac{-15}{23}\)
Mà \(\frac{-15}{25}< \frac{-15}{23}\)
\(\Rightarrow\frac{-15}{25}< \frac{-151515}{232323}.\)