Đáp án + Giải thích các bước giải:
a)
`F` nguyên khi `3x-2\vdtsx+3`
`=>3x+9-11\vdotsx+3`
`=>3(x+3)-11\vdotsx+3`
`=>11\vdotsx+3`
`=>x+3\inƯ(11)={11;-11;1;-1}`
`=>x\in{8;-14;-2;-4}`
b)
`G` nguyên khi `x^2-2x+4\vdotsx+1`
`=>x^2+x-3x-3+7\vdotsx+1`
`=>x(x+1)-3(x+1)+7\vdotsx+1`
`=>(x+1)(x-3)+7\vdotsx+1`
`=>7\vdotsx+1`
`=>x+1\inƯ(7)={7;-7;1;-1}`
`=>x\in{6;-8;0;-2}`