Đáp án:
$\\$
`c,`
`C = (2x+3)/(4x+1) (x \ne (-1)/4)`
Để `C` nguyên
`->2x+3 \vdots 4x+1`
`-> 2 (2x+3) \vdots 4x+1`
`-> 4x + 6 \vdots 4x + 1`
`-> 4x +1 + 5 \vdots 4x+1`
Vì `4x+1 \vdots 4x+1`
`-> 5 \vdots 4x+1`
`-> 4x+1 ∈ Ư (5) = {1;-1;5;-5}`
`-> 4x ∈ {0; -2; 4; -6}`
`-> x ∈ {0;(-2)/4; 1; (-6)/3}`
Vì `x ∈ ZZ`
`-> x ∈ {0;1}` (Thỏa mãn)
vậy `x ∈ {0;1}` để `C` nguyên
`d,`
`D = (3x+2)/(7x+1) (x \ne (-1)/7)`
Để `D` nguyên
`->3x+2 \vdots 7x+1`
`-> 7 (3x+2) \vdots 7x+1`
`->21 x + 14\vdots 7x+1`
`-> 21x + 3 +11 \vdots 7x+1`
`-> 3 (7x+1) + 11 \vdots 7x+1`
Vì `7x+1 \vdots 7x+1 -> 3 (7x+1) \vdots 7x+1`
`->11 \vdots 7x+1`
`-> 7x+1 ∈ Ư (11) = {1;-1;11;-11}`
`-> 7x ∈ {0; -2; 10; -12}`
`-> x ∈ {0; (-2)/7; 10/7; (-12)/7}`
Vì `x ∈ ZZ`
`-> x ∈ {0}` (Thỏa mãn)
Vậy `x ∈ {0}` để `D` nguyên
