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:\(\dfrac{-4}{3};\dfrac{-3}{4};\dfrac{1}{12}\)

a)

\(7\sqrt{2}=\sqrt{49.2}=\sqrt{98}\\ 2\sqrt{8}=\sqrt{4.8}=\sqrt{32}\\ 5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)

Do 98 > 50 > 32 > 28 nên \(\sqrt{98}>\sqrt{50}>\sqrt{32}>\sqrt{28}\)

=> \(7\sqrt{2}>5\sqrt{2}>2\sqrt{8}>\sqrt{28}\)

b)

\(3\sqrt{10}=\sqrt{9.10}=\sqrt{90}\\ 5\sqrt{3}=\sqrt{25.3}=\sqrt{75}\)

\(\dfrac{20}{\sqrt{5}}=\dfrac{20\sqrt{5}}{5}=4\sqrt{5}=\sqrt{16.5}=\sqrt{80}\)

\(12\sqrt{\dfrac{2}{3}}=\sqrt{144.\dfrac{2}{3}}=\sqrt{96}\)

Do 96 > 90 > 80 > 75 => \(\sqrt{96}>\sqrt{90}>\sqrt{80}>\sqrt{75}\)

=> \(12\sqrt{\dfrac{2}{3}}>3\sqrt{10}>\dfrac{20}{\sqrt{5}}>5\sqrt{3}\)

1)                -15 < -1 < 0 < 3 < 5 < 8

2)                2000 > 10 > 4 > 0 > -9 > -97

Hướng dẫn:

a) Sắp xếp các số nguyên theo thứ tự tăng dần là:  − 17 ;    − 7 ;    − 3 ;    0 ;    5 ;    8 ;    100.

b) Sắp xếp các số nguyên theo thứ tự giảm dần là:  17 ;    0 ;    − 1 ;    − 6 ;    − 10 ;    − 31.

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

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

a)-15;-1;0;3;5;8
b)2000;10;4;0;-9;-97

a, -15; -1; 0; 3; 5; 8

b, 2000; 10; 4; 0; -9; -97