Đáp án:

`x\in {1;0;3;-1;-7;-2}`

Giải thích các bước giải:

Điều kiện:`x\in ZZ`

Đặt `A=1-(8x+6)/(x^2+1)`

`\qquad A=(x^2+1-8x-6)/(x^2+1)`

`\qquad A=(x^2-8x-5)/(x^2+1)`

Xét `A+7`

`A+7=(x^2-8x-5)/(x^2+1)+7`

`A+7=(x^2-8x-5+7x^2+7)/(x^2+1)`

`A+7=(8x^2-8x+2)/(x^2+1)`

`A+7=(2(4x^2-4x+1))/(x^2+1)`

`A+7=(2(2x-1)^2)/(x^2+1)`

`x^2+1>=1>0`

Mà `2(2x+1)^2>=0`

`=>A+7>=0`

`=>A>=-7(1)`

Xét `A-3`

`A-3=(x^2-8x-5)/(x^2+1)-3`

`A-3=(x^2-8x-5-3x^2-3)/(x^2+1)`

`A-3=(-2x^2-8x-8)/(x^2+1)`

`A-3=(-2(x^2+4x+4))/(x^2+1)`

`A-3=(-2(x+2)^2)/(x^2+1)`

Vì `x^2+1>=1>0`

Mà `-2(x+2)^2<=0`

`=>A-3<=0`

`=>A<=3(2)`

`(1)(2)=>-7<=A<=3`

Mà `A\inZZ`

`=>A\in{-7;-6;-5;-4;-3;-2;-1;0;1;2;3}`

`**A=-7`

`<=>(x^2-8x-5)/(x^2+1)=-7`

`<=>x^2-8x-5=-7x^2-7`

`<=>8x^2-8x+2=0`

`<=>2(2x-1)^2=0`

`<=>2x-1=0`

`<=>2x=1`

`<=>x=1/2(ktm)`

`**A=-6`

`<=>(x^2-8x-5)/(x^2+1)=-6`

`<=>x^2-8x-5=-6x^2-6`

`<=>7x^2-8x+1=0`

`<=>7x^2-7x-x+1=0`

`<=>7x(x-1)-(x-1)=0`

`<=>(x-1)(7x-1)=0`

`<=>[(x=1(tm)),(x=1/7(ktm)):}`

`<=>x=1`

`**A=-5`

`<=>(x^2-8x-5)/(x^2+1)=-5`

`<=>x^2-8x-5=-5x^2-5`

`<=>6x^2-8x=0`

`<=>3x^2-4x=0`

`<=>x(3x-4)=0`

`<=>[(x=0(tm)),(x=4/3(ktm)):}`

`<=>x=0`

`**A=-4`

`<=>(x^2-8x-5)/(x^2+1)=-4`

`<=>x^2-8x-5=-4x^2-4`

`<=>5x^2-8x-1=0`

`<=>x^2-8/5x-1/5=0`

`<=>x^2-2*x*4/5+16/25=21/25`

`<=>(x-4/5)^2=21/25`

`<=>x-4/5=(+-\sqrt{21})/5`

`<=>x=(+-\sqrt{21}+4)/5`(loại)

`**A=-3`

`<=>(x^2-8x-5)/(x^2+1)=-3`

`<=>x^2-8x-5=-3x^2-3`

`<=>4x^2-8x-2=0`

`<=>2x^2-4x-1=0`

`<=>x^2-2x-1/2=0`

`<=>x^2-2x+1=3/2`

`<=>(x-1)^2=6/4`

`<=>x-1=(+-\sqrt{6})/2`

`<=>x=(+-\sqrt{6}+2)/2`(loại)

`**A=-2`

`<=>(x^2-8x-5)/(x^2+1)=-2`

`<=>x^2-8x-5=-2x^2-2`

`<=>3x^2-8x-3=0`

`<=>3x^2-9x+x-3=0`

`<=>3x(x-3)+x-3=0`

`<=>(x-3)(3x+1)=0`

`<=>[(x=3(tm)),(x=-1/3(ktm)):}`

`<=>x=3`

`**A=-1`

`<=>(x^2-8x-5)/(x^2+1)=-1`

`<=>x^2-8x-5=-x^2-1`

`<=>2x^2-8x-4=0`

`<=>x^2-4x-2=0`

`<=>x^2-4x+4-6=0`

`<=>(x-2)^2=6`

`<=>x-2=+-\sqrt{6}`

`<=>x=+-\sqrt{6}+2`(loại)

`**A=0`

`<=>(x^2-8x-5)/(x^2+1)=0`

`<=>x^2-8x-5=0`

`<=>x^2-8x+16=21`

`<=>(x-4)^2=21`

`<=>x-4=+-\sqrt{21}`

`<=>x=4+-\sqrt{21}`(loại)

`**A=1`

`<=>(x^2-8x-5)/(x^2+1)=1`

`<=>x^2-8x-5=x^2+1`

`<=>8x+6=0`

`<=>4x+3=0`

`<=>4x=-3`

`<=>x=-3/4(ktm)`

`**A=2`

`<=>(x^2-8x-5)/(x^2+1)=2`

`<=>x^2-8x-5=2x^2+2`

`<=>x^2+8x+7=0`

`<=>x^2+x+7x+7=0`

`<=>x(x+1)+7(x+1)=0`

`<=>(x+1)(x+7)=0`

`<=>[(x=-1(tm)),(x=-7(tm)):}`

`**A=3`

`<=>(x^2-8x-5)/(x^2+1)=3`

`<=>x^2-8x-5=3x^2+3`

`<=>2x^2+8x+8=0`

`<=>x^2+4x+4=0`

`<=>(x+2)^2=0`

`<=>x=-2(tm)`

Vậy với `x\in {1;0;3;-1;-7;-2}` thì `1-(8x+6)/(x^2+1)\in ZZ.`