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Đáp án:

Giải thích các bước giải:

Ta có:

`220 = 2^2 . 5 . 11`

`240 = 2^4 . 3 . 5`

`300 = 2^2 . 3 . 5^2`

`-> ƯCLN(220,240,300) = 2^2 . 5 = 20`

`-> ƯC(220,240,300) = Ư(20)`

`-> ƯC(220,240,300) = {±1; ±2; ±4; ±5; ±10; ±20}`

`-> BCNNNN(220,240,300) = 2^4 . 3 . 5^2 . 11 = 13200`

`-> BC(220,240,300) = B(13200)`

`-> BC(220,240,300) = {0; 13200; 26400; 39600; 52800;...}`