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a)\(\dfrac{-5}{6};\dfrac{-3}{4};\dfrac{7}{24};\dfrac{2}{3};\dfrac{7}{8};\dfrac{16}{17}\)
b)\(\dfrac{20}{23};\dfrac{205}{107};\dfrac{214}{315};\dfrac{7}{10};\dfrac{-5}{8};\dfrac{-16}{19}\)
-5/8=-15/24=-30/48
-9/16=-27/48
-2/3=-32/48
-7/12=-28/48
=>-2/3<-5/8<-7/12<-9/16
a) \(\dfrac{-5}{6}=\dfrac{-340}{408}\);\(\dfrac{7}{8}=\dfrac{357}{408}\);\(\dfrac{7}{24}=\dfrac{119}{408}\)
\(\dfrac{16}{17}=\dfrac{384}{408}\); \(\dfrac{-3}{4}=\dfrac{-306}{408}\); \(\dfrac{2}{3}=\dfrac{272}{408}\)
Do đó: \(\dfrac{-5}{6}< \dfrac{-3}{4}< \dfrac{7}{24}< \dfrac{2}{3}< \dfrac{7}{8}< \dfrac{16}{17}\)
a) ta có: \(\frac{-5}{6}=\frac{-10}{12};\frac{-3}{4}=\frac{-9}{12}\)
\(\frac{7}{8}=\frac{357}{408};\frac{7}{24}=\frac{119}{408};\frac{16}{17}=\frac{384}{408};\frac{2}{3}=\frac{272}{408}\)
\(\Rightarrow\frac{-10}{12}< \frac{-9}{12}< \frac{119}{408}< \frac{272}{408}< \frac{357}{408}< \frac{384}{408}\)
\(\Rightarrow\frac{-5}{6}< \frac{-3}{4}< \frac{7}{24}< \frac{2}{3}< \frac{7}{8}< \frac{16}{17}\)
b) ta có: \(\frac{205}{107}=1\frac{98}{107}>1\)\(\frac{-5}{8};\frac{7}{10};\frac{-16}{19};\frac{20}{26}=\frac{10}{13};\frac{214}{315}< 1\)
ta có: \(\frac{-5}{8}=\frac{-95}{152};\frac{-16}{19}=\frac{-128}{152}\)
\(\frac{7}{10}=\frac{5733}{8190};\frac{20}{26}=\frac{10}{13}=\frac{6300}{8190};\frac{214}{315}=\frac{5564}{8190}\)
\(\Rightarrow\frac{205}{107}>\frac{6300}{8190}>\frac{5733}{8190}>\frac{5564}{8190}>\frac{-95}{152}>\frac{-128}{152}\)
=> ...
Bài 1:
\(a.-5;-3;-2;0;1;2;4\)
\(b.-36;-8;-6;-5;-4;0;6;8;12;15\)
\(c.-129;-98;0;3;27;35\)
Bài 2:
\(a.15;14;9;0;-3;-7;-16\)
\(b.100;17;5;0;-1;-2;-3;-13;-99\)
\(\cdot DuyNam\)
`a, -5/6; -3/4; 7/24; 7/8; 16/17`
`b, 205/107; 20/23; 214/315; -5/8; -16/19`
a) -97 , -16 , -2 , 0 , 5 , 18
b) I-2018I , 100 , 12 , 0 ,-15 , -40 ,-100
CHÚC BẠN HỌC TỐT
Thứ tự tăng dần : \(\dfrac{3}{4},\dfrac{5}{8},\dfrac{7}{8},\dfrac{15}{16},\dfrac{18}{16}\)
ta có; 5/8 = 10/16
7/8 = 14/16
5/8 < 13/16 < 7/8 < 15/16 < 18/16