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a: Ta có: -97<-72<-36<-15<-7<-2
=>\(\frac{-97}{24}<\frac{-72}{24}<\frac{-36}{24}<-\frac{15}{24}<-\frac{7}{24}<-\frac{2}{24}\)
b: Ta có: \(\frac{-3}{5}=\frac{-3\cdot4}{5\cdot4}=\frac{-12}{20}\)
\(\frac{1}{-5}=\frac{-1}{5}=\frac{-1\cdot4}{5\cdot4}=\frac{-4}{20}\)
\(\frac{-3}{4}=\frac{-3\cdot5}{4\cdot5}=\frac{-15}{20}\)
mà -4>-12>-15
nên \(\frac{1}{-5}>\frac{-3}{5}>-\frac34\)
=>\(\frac{1}{-5}>\frac{-3}{5}>\frac{-3}{4}>-1\left(1\right)\)
Ta có: \(-1=\frac{-12}{12};\frac{5}{-3}=\frac{-5}{3}=\frac{-5\cdot4}{3\cdot4}=\frac{-20}{12};\frac{5}{-4}=\frac{-5}{4}=\frac{-5\cdot3}{4\cdot3}=\frac{-15}{12}\)
mà -12>-15>-20
nên \(-1>\frac{-5}{4}>-\frac53\) (2)
Từ (1),(2) suy ra \(\frac{1}{-5}>\frac{-3}{5}>\frac{-3}{4}>\frac{-5}{4}>-\frac53\)
\(0>\frac{-9}{10}>\frac{-13}{15}>\frac{4}{-12}>\frac{11}{-6}\)
CHÚC BN HỌC TỐT!
\(a\)) \(-17;-2;0;1;2;5\)
\(b\))\(2001;15;7;0;-8;-101\)
Giải:
a)
\(\dfrac{7}{48}=\dfrac{105}{720};\)
\(\dfrac{11}{72}=\dfrac{110}{720};\)
\(\dfrac{17}{120}=\dfrac{102}{720}\)
Vì \(102< 105< 110\)
\(\Leftrightarrow\dfrac{102}{720}< \dfrac{105}{720}< \dfrac{110}{720}\)
\(\Leftrightarrow\dfrac{17}{120}< \dfrac{7}{48}< \dfrac{11}{72}\)
Vậy ...
b) \(\dfrac{31}{49}=\dfrac{60140}{95060};\)
\(\dfrac{62}{97}=\dfrac{60760}{95060};\)
\(\dfrac{93}{140}=\dfrac{63147}{95060}\)
Vì \(60140< 60760< 63147\)
\(\Leftrightarrow\dfrac{60140}{95060}< \dfrac{60760}{95060}< \dfrac{63147}{95060}\)
\(\Leftrightarrow\dfrac{31}{49}< \dfrac{62}{97}< \dfrac{93}{140}\)
Vậy ...
a ) \(\dfrac{7}{48}\) = \(\dfrac{105}{720}\)
\(\dfrac{11}{72}\) = \(\dfrac{110}{720}\)
\(\dfrac{17}{120}\) = \(\dfrac{102}{720}\)
Vì 102 < 105 < 110
\(\Leftrightarrow\) \(\dfrac{102}{720}\) < \(\dfrac{105}{720}\) < \(\dfrac{110}{720}\)
\(\Leftrightarrow\) \(\dfrac{17}{120}\) < \(\dfrac{7}{48}\) < \(\dfrac{11}{72}\)
Vậy .....................
( k cho tớ nha . Tớ chỉ bt lm phần a )
\(\frac{-5}{6}< \frac{-3}{4}< \frac{7}{24}< \frac{2}{3}< \frac{7}{8}\)
CHÚC BN HỌC TỐT!!!!
a ) u K v ^ < D E F ^ .
b ) z M x ^ < D E F ^
c ) u K v ^ < z M x ^ < D E F ^