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Ta có: \(\dfrac{5}{7} = \dfrac{{5.4}}{{7.4}} = \dfrac{{20}}{{28}}\) và \(\dfrac{{ - 3}}{4} = \dfrac{{ - 3.7}}{{4.7}} = \dfrac{{ - 21}}{{28}}\)
Như vậy, \(\dfrac{{20}}{{28}} + \dfrac{{ - 21}}{{28}} = \dfrac{{20 + \left( { - 21} \right)}}{{28}} = \dfrac{-1}{{28}}\)
a ) − 1 7 ; b ) − 13 15 ; c ) − 19 24
d ) − 19 24 ; e ) − 37 90 ; f ) − 41 11
a) \(\frac{8}{2^3.25}=\frac{8}{8.25}=\frac{1}{25}\)
\(\frac{10}{2^2.5^2.7}=\frac{100}{10^2.7}=\frac{1}{70}\)
Mẫu chung của 2 phân số là 350
1: B là số nguyên
=>n-3 thuộc {1;-1;5;-5}
=>n thuộc {4;2;8;-2}
3:
a: -72/90=-4/5
b: 25*11/22*35
\(=\dfrac{25}{35}\cdot\dfrac{11}{22}=\dfrac{5}{7}\cdot\dfrac{1}{2}=\dfrac{5}{14}\)
c: \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}=\dfrac{54-34}{189-119}=\dfrac{20}{70}=\dfrac{2}{7}\)
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 \(-\dfrac{138}{150}< -\dfrac{105}{150}< -\dfrac{60}{150}< -\dfrac{51}{150}< -\dfrac{37}{150}\)
=>\(\dfrac{23}{-25}< \dfrac{-7}{10}< \dfrac{-2}{5}< \dfrac{-17}{50}< \dfrac{37}{-150}\)
