Đáp án:
`1,` Gọi ƯCLN(n+1;2n+3)=d
Ta có: `n+1`$\vdots$`d`; `2n+3` $\vdots$ `d`
`=> 2n+2`$\vdots$`d`; `2n+3` $\vdots$ `d`
`=> 2n+3-(2n+2)`$\vdots$`d`
`=> 1`$\vdots$`d`
`=> d∈Ư(1)`
`=> d=1`
`=> đpcm`
`2, Gọi ƯCLN(n-1;2n)=d
Ta có: `n-1`$\vdots$ `d`; `2n`$\vdots$ `d`
`=> 2n-2`$\vdots$ `d`; 2n`$\vdots$ `d`
`=> 2n-(2n-2)`$\vdots$ `d`
`=> 2`$\vdots$ `d`
`=> d∈Ư(2)={1;2}`
`=>`\(\left[ \begin{array}{l}d=1\\d=2\end{array} \right.\) (xem lại đề)
`3,` Gọi ƯCLN(2n+1;3n+1)=d
Ta có: `2n+1`$\vdots$ `d`; `3n+1`$\vdots$ `d`
`=> 3(2n+1)`$\vdots$ `d`; `2(3n+1)`$\vdots$ `d`
`=> 6n+3`$\vdots$ `d`;` 6n+2`$\vdots$ `d`
`=> 6n+3-(6n+2)`$\vdots$ `d`
`=> 1`$\vdots$ `d`
`=> d∈Ư(1)`
`=> d=1`
`=> đpcm`
`4,`Gọi ƯCLN(2n+3;4n+5)=d
Ta có: `2n+3`$\vdots$ `d`; `4n+5`$\vdots$ `d`
`=> 2(2n+3)`$\vdots$ `d`; `4n+5`$\vdots$ `d`
`=> 4n+6`$\vdots$ `d`; `4n+5`$\vdots$ `d`
`=> 4n+6-(4n+5)`$\vdots$ `d`
`=> 1`$\vdots$ `d`
`=> d∈Ư(1)`
`=> d=1`
`=> đpcm`
Giải thích các bước giải:
