bài 1:Tìm số nguyên x 25-(25-x)=-8+3(x-4) 5+|x+3|+9 3-(7+x)=-16+2.(5-x) x-(-25+x)=13-x 15-(30+x)=x-(27-|-8|) bài 2:Tìm số nguyên n để 2n+3 chia hết c

`B1:`

`25-(25-x)=-8+3(x-4)`

`=>25-25+x=-8+3x-12`

`=>x=3x-20`

`=>2x=20`

`=>x=10`

Vậy `x=10` thì `25-(25-x)=-8+3(x-4)`.

`5+|x+3|=9`

`=>|x+3|=9-5=4`

`=>` \(\left[ \begin{array}{l}x+3=4\\x+3=-4\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x=1\\x=-7\end{array} \right.\) 

Vậy `x=1` hoặc `x=-7` thì `5+|x+3|=9`.

`3-(7+x)=-16+2(5-x)`

`=>3-7-x=-16+10-2x`

`=>-4-x=-2x-6`

`=>x=-2`

Vậy `x=-2` thì `3-(7+x)=-17+2(5-x)`.

`x-(-25+x)=13-x`

`=>x+25-x=13-x`

`=>25=13-x`

`=>x=13-25=-12`

Vậy `x=-12` thì `x-(-25+x)=13-x`.

`15-(30+x)=x-(27-|-8|)`

`=>15-30-x=x-(27-8)`

`=>-15-x=x-19`

`=>2x=4`

`=>x=2`

Vậy `x=2` thì `15-(30+x)=x-(27-|-8|)`.

`B2:`

`2n+3 vdots n-3(n in ZZ)`

`=>2n-6+9 vdots n-3`

`=>2(n-3)+9 vdots n-3`

`=>9 vdots n-3`(do `2(n-3) vdots n-3`)

`=>n-3 in Ư(9)={+-1,+-3,+-9}`

Ta có bảng sau:

$\begin{array}{|c|c|c|}\hline n-3&1&-1&3&-3&9&-9 \\\hline  n&4&2&6&0&12&-6 \\\hline KL&TM&TM&TM&TM&TM&TM \\\hline\end{array}$

Vậy với `n in {4,2,6,0,12,-6}` thì `2n+3 vdots n-3`.

Trả lời