Đáp án:
`a)x\in \{-15;-5;-3;7\}`
`b)x\in\{-1;0\}`
`c)x\in \{-5;-4;-2;-1\}`
`d)x\in \{-10;2;4;16\}`
`e)x\in \{-2;0\}`
Giải thích các bước giải:
`a)` Để `\frac{x-7}{x+4}\in Z`
`=>x-7\vdots x+4`
`=>(x+4)-11\vdots x+4`
Do `x+4\vdots x+4`
`=>11\vdots x+4`
`=>x+4\in Ư(11)=\{-11;-1;1;11\}`
`=>x\in \{-15;-5;-3;7\}`
`b)` Để `\frac{6x+4}{3x+1}\in Z`
`=>6x+4\vdots 3x+1`
`=>6x+2+2\vdots 3x+1`
`=>2(3x+1)+2\vdots 3x+1`
Do `2(3x+1)\vdots 3x+1`
`=>2\vdots 3x+1`
`=>3x+1\in Ư(2)=\{-2;-1;1;2\}`
`=>3x\in \{-3;-2;0;1\}`
`=>x\in \{-1;0\}`
`c)` Để `\frac{x+5}{x+3}\in Z`
`=>x+5\vdots x+3`
`=>(x+3)+2\vdots x+3`
Do `x+3\vdots x+3`
`=>2\vdots x+3`
`=>x+3\in Ư(2)=\{-2;-1;1;2\}`
`=>x\in \{-5;-4;-2;-1\}`
`d)` Để `\frac{2x+7}{x-3}\in Z`
`=>2x+7\vdots x-3`
`=>2x-6+13\vdots x-3`
`=>2(x-3)+13\vdots x-3`
Do `2(x-3)\vdots x-3`
`=>13\vdots x-3`
`=>x-3\in Ư(13)=\{-13;-1;1;13\}`
`=>x\in \{-10;2;4;16\}`
`e)` Để `\frac{6x-3}{3x+1}\in Z`
`=>6x-3\vdots 3x+1`
`=>6x+2-5\vdots 3x+1`
`=>2(3x+1)-5\vdots 3x+1`
Do `2(3x+1)\vdots 3x+1`
`=>5\vdots 3x+1`
`=>3x+1\in Ư(5)=\{-5;-1;1;5\}`
`=>3x\in \{-6;-2;0;4\}`
`=>x\in \{-2;0\}`
