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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 \(-\dfrac{138}{150}< -\dfrac{105}{150}< -\dfrac{60}{150}< -\dfrac{51}{150}< -\dfrac{37}{150}\)
=>\(\dfrac{23}{-25}< \dfrac{-7}{10}< \dfrac{-2}{5}< \dfrac{-17}{50}< \dfrac{37}{-150}\)
1. a, \(\frac{6}{7}\)=\(\frac{60}{70}\);\(\frac{11}{10}\)=\(\frac{77}{70}\)
vì \(\frac{60}{70}\)<\(\frac{77}{70}\)nên \(\frac{6}{7}\)<\(\frac{11}{10}\)
b, \(\frac{-5}{17}\)<0<\(\frac{2}{7}\)
c, \(\frac{419}{-723}\)<0<\(\frac{-697}{-313}\)
2.
Ta có :\(\frac{2}{6}\)=\(\frac{20}{60}\);\(\frac{5}{12}\)=\(\frac{25}{60}\);\(\frac{4}{15}\)=\(\frac{16}{60}\);\(\frac{8}{20}\)=\(\frac{24}{60}\);\(\frac{10}{30}\)=\(\frac{20}{60}\)
Vì \(\frac{16}{60}\)<\(\frac{20}{60}\)<\(\frac{24}{60}\)<\(\frac{25}{60}\)nên \(\frac{4}{15}\)<\(\frac{2}{6}\)=\(\frac{10}{30}\)<\(\frac{8}{20}\)<\(\frac{5}{12}\)
\(-5,\dfrac{5}{-7},\dfrac{-3}{4},0,\dfrac{-5}{-6},\dfrac{4}{5},\dfrac{39}{7}\)
a: Số đối của 25 là -25
Số đối của -7 là 7
Số đối của 0 là 0
b: -2018<-5<-3<0<3<7
B1:
a) \(\frac{9}{10}>\frac{9}{42}>\frac{5}{42}\)
b) \(-\frac{4}{27}< \frac{10}{-73}\)
B2: \(\frac{5}{-6};-\frac{7}{12};\frac{5}{8};\frac{3}{4}\)