a)sắp xếp theo thứ tự tăng dần là: \(\dfrac{-1}{2};\dfrac{2}{7};\dfrac{2}{5}\)

b)sắp xếp theo thứ tự tăng dần là: \(\dfrac{-11}{4};\dfrac{-7}{3};\dfrac{12}{5}\)

c)sắp xếp theo thứ tự tăng dần là:\(\dfrac{-18}{5};\dfrac{9}{-2};\dfrac{10}{3}\)

d)\(sửa\dfrac{3-}{4}=\dfrac{-3}{4}\)

sắp xếp theo thứ tự tăng dần là:\(\dfrac{-4}{3};\dfrac{-3}{4};\dfrac{1}{12}\)

\(sin16^0=sin\left(90^0-74^0\right)=cos74^0\)

\(sin60^0=sin\left(90^0-60^0\right)=cos30^0\)

Sắp xếp: \(cos16^0,cos30^0,cos43^0,cos52^0,cos74^0\)

cos16=sin 74

cos43=sin47

cos52=sin38

Vì 16<38<47<60<74

nên sin 16<sin 38<sin 47<sin60<sin74

=>sin 16<cos52<cos43<sin60<cos16

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

Thứ tự tăng dần: \(\dfrac{3}{4};\dfrac{5}{8};\dfrac{7}{8};\dfrac{15}{16};\dfrac{18}{16}\)

a) -97 , -16 , -2 , 0 , 5 , 18

b)  I-2018I , 100 , 12 , 0 ,-15 , -40 ,-100

CHÚC BẠN HỌC TỐT

Bài 7:

7.1: I là trung điểm của AB

=>\(AB=2\cdot IA=4\left(cm\right)\)

7.2:

C nằm giữa A và B

=>AC+CB=AB

=>CB=10-8=2(cm)

C là trung điểm của NB

=>NC=CB=2cm

C là trung điểm của NB

=>\(NB=2\cdot NC=2\cdot2=4\left(cm\right)\)

Bài 6:

a: \(\dfrac{4}{5}=\dfrac{4\cdot6}{5\cdot6}=\dfrac{24}{30}\)

\(\dfrac{8}{15}=\dfrac{8\cdot2}{15\cdot2}=\dfrac{16}{30}\)

\(-\dfrac{3}{2}=\dfrac{-3\cdot15}{2\cdot15}=-\dfrac{45}{30}\)

b: \(2=\dfrac{2\cdot45}{45}=\dfrac{90}{45}\)

\(\dfrac{-10}{5}=\dfrac{-10\cdot9}{5\cdot9}=\dfrac{-90}{45}\)

\(\dfrac{7}{-9}=\dfrac{-7}{9}=\dfrac{-7\cdot5}{9\cdot5}=\dfrac{-35}{45}\)

c: \(\dfrac{3}{-2}=\dfrac{-3}{2}=\dfrac{-3\cdot6}{2\cdot6}=\dfrac{-18}{12}\)

\(\dfrac{5}{-6}=\dfrac{-5}{6}=\dfrac{-5\cdot2}{6\cdot2}=\dfrac{-10}{12}\)

\(\dfrac{-6}{4}=\dfrac{-6\cdot3}{4\cdot3}=\dfrac{-18}{12}\)

d: \(-\dfrac{1}{2}=\dfrac{-1\cdot15}{2\cdot15}=\dfrac{-15}{30}\)

\(\dfrac{4}{3}=\dfrac{4\cdot10}{3\cdot10}=\dfrac{40}{30}\)

\(\dfrac{6}{-5}=\dfrac{-6}{5}=\dfrac{-6\cdot6}{5\cdot6}=\dfrac{-36}{30}\)

bài 5:

a: \(\dfrac{3}{4}=\dfrac{9}{12};\dfrac{-3}{12}=\dfrac{-3}{12};\dfrac{-2}{3}=-\dfrac{8}{12};\dfrac{-1}{-6}=\dfrac{1}{6}=\dfrac{2}{12}\)

mà -8<-3<2<9

nên \(-\dfrac{8}{12}< -\dfrac{3}{12}< \dfrac{2}{12}< \dfrac{9}{12}\)

=>\(\dfrac{-2}{3}< \dfrac{-3}{12}< \dfrac{-1}{-6}< \dfrac{3}{4}\)

b: Ta có: \(\dfrac{-7}{9}=\dfrac{-28}{36};\dfrac{-1}{3}=\dfrac{-12}{36};-1=-\dfrac{36}{36}\)

mà -36<-28<-12

nên \(-1< -\dfrac{28}{36}< -\dfrac{12}{36}\)

=>\(-1< \dfrac{-7}{9}< -\dfrac{1}{3}< 0\)

\(\dfrac{5}{12}=\dfrac{15}{36};\dfrac{-1}{-4}=\dfrac{1}{4}=\dfrac{9}{36}\)

mà 9<15

nên \(0< \dfrac{1}{4}< \dfrac{5}{12}\)

=>\(-1< -\dfrac{7}{9}< -\dfrac{1}{3}< 0< \dfrac{1}{4}< \dfrac{5}{12}\)

c: \(\dfrac{-1}{-2};0;\dfrac{3}{10};1;\dfrac{-2}{-5};\dfrac{3}{-4}\)

\(-\dfrac{3}{4}< 0\)

\(\dfrac{-1}{-2}=\dfrac{1}{2}=\dfrac{5}{10};\dfrac{3}{10}=\dfrac{3}{10};1=\dfrac{10}{10};\dfrac{-2}{-5}=\dfrac{4}{10}\)

mà 3<4<5<10

nên \(\dfrac{3}{10}< \dfrac{4}{10}< \dfrac{5}{10}< \dfrac{10}{10}\)

=>\(0< \dfrac{3}{10}< \dfrac{-2}{-5}< \dfrac{-1}{-2}< 1\)

=>\(-\dfrac{3}{4}< 0< \dfrac{3}{10}< \dfrac{-2}{-5}< \dfrac{-1}{-2}< 1\)

d: \(-\dfrac{37}{150}=\dfrac{-37}{150};\dfrac{17}{-50}=\dfrac{-17}{50}=\dfrac{-51}{150}\)

\(\dfrac{23}{-25}=\dfrac{-23}{25}=\dfrac{-138}{150};\dfrac{-7}{10}=\dfrac{-105}{150};\dfrac{-2}{5}=-\dfrac{60}{150}\)

mà -138<-105<-60<-51<-37

nên \(-\dfrac{138}{150}< -\dfrac{105}{150}< -\dfrac{60}{150}< -\dfrac{51}{150}< -\dfrac{37}{150}\)

=>\(\dfrac{23}{-25}< \dfrac{-7}{10}< \dfrac{-2}{5}< \dfrac{-17}{50}< \dfrac{37}{-150}\)

-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)\(\dfrac{8}{15}< \dfrac{11}{15}< \dfrac{15}{15}< \dfrac{16}{15}< \dfrac{17}{15}< \dfrac{19}{15}\)

b)\(\dfrac{19}{42}< \dfrac{19}{35}< \dfrac{19}{21}< \dfrac{19}{19}< \dfrac{19}{17}\)

c)\(\dfrac{8}{10}< \dfrac{6}{7}< \dfrac{27}{25}< \dfrac{16}{14}\)

a) 8/15 ;  11/15 ; 15/15 ; 16/15 ; 17/15 ; 19/15

b) 19/42 19/35 ; 19/21 ; 19/19 ; 19/17

c) 8/10 ; 6/7 ; 27/25 ; 16/14

a)-18,-7/9.-2/9,11

b)-17/40;-4/15,-2/5,17