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\(\dfrac{12+24+72}{18+36+108}=\dfrac{108}{162}=\dfrac{2}{3}=\dfrac{12}{18}=\dfrac{36}{54}\)
\(\dfrac{12-24+72}{18-36+108}=\dfrac{60}{90}=\dfrac{2}{3}=\dfrac{12}{18}=\dfrac{36}{54}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{12}{18}=\dfrac{24}{36}=\dfrac{72}{108}=\dfrac{12+24+72}{18+36+108}=\dfrac{12-24+72}{18-36+108}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
c, Gọi ƯCLN(a; b) = d; d \(\in\) k
⇒ d = 1944 : 108 = 18
⇒ a = 18.k; b = 18.n (k;n) =1; k;n \(\in\) N*
⇒18.k.18.n = 1944
⇒k.n =1944 : (18.18)
k.n = 6
6 = 2.3 Ư(6) = {1; 2; 3;6)
⇒(k; n) = (1; 6); (2; 3); (3; 2); (6; 1)
⇒ (a; b) = (18; 108); (36; 54); (54; 36); (108; 18)
Vì a> b nên (a; b) = (54; 36); (108; 18)
a, a + b = 72; Ư CLN(a; b) = 9 (a > b)
a = 9.k; b = 9.d (k; d) = 1; k; d \(\in\) N*; k >d
9.k + 9.d = 72
9.(k + d) = 72
k + d = 72 : 9
k + d = 8
(k; d) =(1; 7); (2; 6); (3; 5); (4; 4); (5; 3); (6; 2); (7; 1)
vì (k;d) = 1; k > d ⇒ (k;d) = (5; 3); (7; 1)
⇒ (a; b) = (45; 27); (63; 9)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\dfrac{-11}{12}và\dfrac{17}{-18}\) \(\Leftrightarrow\dfrac{-11}{12}và\dfrac{-17}{18}\) \(\Leftrightarrow\dfrac{-33}{36}và\dfrac{-34}{36}\)
Ta thấy rằng : \(-33>-34\Rightarrow\dfrac{-33}{36}>\dfrac{-34}{36}\)
Hay : \(\dfrac{-11}{12}>\dfrac{17}{-18}\)
\(b)\dfrac{-14}{-21}và\dfrac{-60}{-72}\)
Ta có : \(\dfrac{-14}{-21}\text{=}\dfrac{-14:-7}{-21:-7}\text{=}\dfrac{2}{3}\text{=}\dfrac{4}{6}\)
\(\dfrac{-60}{-72}\text{=}\dfrac{-60:-12}{-72:-12}=\dfrac{5}{6}\)
Do đó : \(\dfrac{-14}{-21}< \dfrac{-60}{-72}\)
\(c)\dfrac{2135}{13790}và\dfrac{4}{3}\)
Xét phân số : \(\dfrac{2135}{13790}\) ta thấy rằng : \(tử< mẫu\left(2135< 13790\right)\)
\(\Rightarrow\dfrac{2135}{13790}< 1\)
Xét phân số : \(\dfrac{4}{3}có\) : \(tử>mẫu\left(4>3\right)\)
\(\Rightarrow\dfrac{4}{3}>1\)
Do đó : \(\dfrac{2135}{13790}< \dfrac{4}{3}\)
\(d)\dfrac{2022}{2021}và\dfrac{10}{9}\)
Ta thấy rằng : \(\dfrac{2022}{2021}-\dfrac{1}{2021}\text{=}1\)
\(\dfrac{10}{9}-\dfrac{1}{9}\text{=}1\)
Mà : \(\dfrac{1}{9}>\dfrac{1}{2021}\)
\(\Rightarrow\dfrac{2022}{2021}< \dfrac{10}{9}\)
\(e)\dfrac{35}{36}và\dfrac{16}{17}\)
Ta có : \(\dfrac{35}{36}+\dfrac{1}{36}\text{=}1\)
\(\dfrac{16}{17}+\dfrac{1}{17}\text{=}1\)
Mà : \(\dfrac{1}{36}< \dfrac{1}{17}\)
\(\Rightarrow\dfrac{35}{36}>\dfrac{16}{17}\)
\(f)-1,3< -1,2\)
a) Ta có:
\(-\dfrac{11}{12}=\dfrac{1}{12}-1\)
\(-\dfrac{17}{18}=\dfrac{1}{18}-1\)
Mà: \(\dfrac{1}{12}>\dfrac{1}{18}\)
Hay: \(\dfrac{1}{12}-1>\dfrac{1}{18}-1\Rightarrow-\dfrac{11}{12}>-\dfrac{17}{18}\)
b) Ta có:
\(\dfrac{-14}{-21}=\dfrac{2}{3}=\dfrac{4}{6}\)
\(\dfrac{-60}{-72}=\dfrac{5}{6}\)
Mà: \(5>4\Rightarrow\dfrac{-60}{-72}>\dfrac{-14}{-21}\)
c) Ta có:
\(\dfrac{2135}{13790}=\dfrac{61}{394}< 1\) (tử nhỏ hơn mẫu)
\(\dfrac{4}{3}>1\) (tử lớn hơn mẫu)
Ta có: \(\dfrac{61}{394}< \dfrac{4}{3}\Rightarrow\dfrac{2135}{13790}< \dfrac{4}{3}\)
d) Ta có:
\(\dfrac{2022}{2021}=\dfrac{1}{2021}+1\)
\(\dfrac{10}{9}=\dfrac{1}{9}+1\)
Ta thấy: \(\dfrac{1}{2021}< \dfrac{1}{9}\Rightarrow\dfrac{1}{2021}+1< \dfrac{1}{9}+1\)
Hay \(\dfrac{2022}{2021}< \dfrac{10}{9}\)
e) Ta có:
\(\dfrac{35}{36}=1-\dfrac{1}{36}\)
\(\dfrac{16}{17}=1-\dfrac{1}{17}\)
Ta có: \(\dfrac{1}{36}< \dfrac{1}{17}\Rightarrow1-\dfrac{1}{36}>1-\dfrac{1}{17}\)
Hay \(\dfrac{35}{36}>\dfrac{16}{17}\)
f) Ta có: \(1,3>1,2\)
\(\Rightarrow-1,3< -1,2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a}{b}=\frac{18}{27}=\frac{2}{3}\)
=> a = 2c ; b = 3c ( c \(\in\)N* và c là số nguyên tố )
Mà ƯCLN( a;b ) = 17 nên ƯCLN( 2c;3c ) = 17 => 2c chia hết cho 17 ; 3c chia hết cho 17
=> 3c - 2 c = c chia hết cho 17
Từ đó suy ra : a = 17 x 2 = 34
b = 17 x 3 = 51
Vậy phân số \(\frac{a}{b}=\frac{34}{51}\)
Mik ko trả lời câu hỏi của bn nữa đâu!
ài 3:
a) Ta có: 18=2⋅3218=2⋅32
24=23⋅324=23⋅3
30=2⋅3⋅530=2⋅3⋅5
Do đó: ƯCLN(18;24;30)=2⋅3=6ƯCLN(18;24;30)=2⋅3=6 và BCNN(18;24;30)=23⋅32⋅5=360BCNN(18;24;30)=23⋅32⋅5=360
b) Ta có: 40=23⋅540=23⋅5
75=3⋅5275=3⋅52
105=3⋅5⋅7105=3⋅5⋅7
Do đó: ƯCLN(40;75;105)=51=5ƯCLN(40;75;105)=51=5 và BCNN(40;75;105)=23⋅3⋅52⋅7=4200BCNN(40;75;105)=23⋅3⋅52⋅7=4200
c) Ta có: 18=2⋅3218=2⋅32
36=22⋅3236=22⋅32
72=23⋅3272=23⋅32
Do đó: ƯCLN(18;36;72)=2⋅32=2⋅9=18ƯCLN(18;36;72)=2⋅32=2⋅9=18 và BCNN(18;36;72)=23⋅32=72BCNN(18;36;72)=23⋅32=72
Bài 4:
a) Ta có: Ư(24)={1;2;3;4;6;8;12;24}
Ư(36)={1;2;3;4;6;9;12;18;36}
Ư(60)={1;2;3;4;5;6;10;12;15;20;30;60}
Do đó: ƯC(24;36;60)={1;2;3;4;6;12}
b) Ta có: B(12)={0;12;24;36;48;60;72;84;96;108;120;...;180;....}
B(30)={0;30;60;90;120;150;180;...}
B(45)={0;45;90;135;180;...}
Do đó: BC(12;30;45)={0;180;360;540;...}