Đáp án:
`a)` Gọi ƯCLN(`n+1;3n+4`)=d
Ta có: `n+1`$\vdots$ `d`; `3n+4`$\vdots$ `d`
`=> 3(n+1)`$\vdots$ `d`; `3n+4`$\vdots$ `d`
`=> 3n+4-(3n+3)`$\vdots$ `d`
`=> 1`$\vdots$ `d`
`=>d∈Ư(1)`
`=> d=1`
`=> đpcm`
`b)` Gọi ƯCLN(`3n+10; 2n+7`)=d
Ta có: `3n+10`$\vdots$ `d`; `2n+7`$\vdots$ `d`
`=> 2(3n+10)`$\vdots$ `d`;`3(2n+7)`$\vdots$ `d`
`=> 6n+20`$\vdots$ `d`;`6n+21`$\vdots$ `d`
`=> 6n+21-(6n+20)`$\vdots$ `d`
`=>d∈Ư(1)`
`=> d=1`
`=> đpcm`
Giải thích các bước giải:
