Đáp án:

Giải thích các bước giải:

Bài `1:`

`a) 60 = 2^2 . 3 . 5`

`132 = 2^2 . 3 . 11`

`=> ƯCLN(60,132) = 2^2 . 3 = 12`

`=> ƯC(60,132) = {±1; ±2; ±3; ±4; ±6; ±12}`

`b) 168 = 2^3 . 3 . 7`

`120 = 2^3 . 3 . 5`

`144 = 2^4 . 3^2`

`=> ƯCLN(168, 120, 144) = 2^3 . 3 = 24`

`=> ƯC(168,120,144) = {±1; ±2; ±3; ±4; ±6; ±8; ±12; ±24}`

Bài `2:`

`612 vdots a; 680 vdots a`

`=> a in ƯC(612,680)`

`612 = 2^2 . 3^2 . 17`

`680 = 2^3 . 5 . 17`

`=> ƯCLN(612,680) = 2^2 . 17 = 68`

`=> ƯC(612,680) = {±1; ±2; ±4; ±17; ±34; ±68}`

Mà `a > 30 => a = {34; 68}`

Bài 1:

a) Ta có:

`60 = 2^2 .3.5`

`132=2^2 .3.11`

`=> text{ƯCLN} (60; 132) = 2^2 . 3 = 12`

`=> text{ƯC} (60 ; 132) = Ư (12) = {±12; ±6; ±4; ±3; ±2; ±1}`

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b) Ta có:

`168 = 2^3 . 3 . 7`

`120 = 2^3 . 3 . 5`

`144 = 2^4 . 3^2`

`=> text{ƯCLN} (168; 120; 144) = 2^3 . 3= 24` 

`=> text{ƯC} (168; 120; 144) = Ư (24) = {±24; ±12; ±8; ±6; ±4; ±3; ±2; ±1}`

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Bài 2:

Theo đề, ta có:

`a in text{ƯC} (612; 680)`

Mà: 

`612 = 2^2 . 3^2 . 17`

`680 = 2^3 . 5 . 17`

`=> text{ƯCLN} (612; 680) = 2^2 . 17 = 68`

`=> a in text{ƯC} (612; 680) = Ư (68) = {±68; ±34; ±17; ±4; ±2; ±1}`

Mà `a>30`

`=> a in {34; 68}`