Đáp án + Giải thích các bước giải:
`a//|2x-1|=5`
`=>2x-1=±5`
`⇒` \(\left[ \begin{array}{l}2x=5+1\\2x=-5+1\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}2x=6\\2x=-4\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
Vậy `x∈{3;-2}`
`b//2x+7` $\vdots$ `x`
`=>7` $\vdots$ `x` . Do `2x` $\vdots$ `x`
`=>x∈Ư(7)={±1;±7}`
Vậy `x∈{±1;±7}`
`c//3x+2` $\vdots$ `2x+1`
`=>6x+4` $\vdots$ `2x+1`
`⇒3(2x+1)+1` $\vdots$ `2x+1`
`⇒1` $\vdots$ `2x+1`
`⇒2x+1∈Ư(1)={±1}`
`⇒2x∈{0;-2}`
`=>x∈{0;-1}`
`d//(x+2).(3-y)=11`
`=>(x+2).(3-y)=11=1.11=(-1).(-11)` . Do `x;y∈Z`
Lập bảng , ta có :
$\begin{array}{|c|c|}\hline x+2&1&11&-1&-11\\\hline 3-y&11&1&-11&-1\\\hline\end{array}$
`⇒`
$\begin{array}{|c|c|}\hline x&-1&9&-3&-13\\\hline y&-8&2&14&4\\\hline\end{array}$
Vậy `(x;y)=(-1;-8);(9;2);(-3;14);(-13;4)`
