Đáp án:
Giải thích các bước giải:
Tìm $x^{}$ nguyên để phân thức sau nguyên: $\frac{x+2}{x^{2}+4}$
Để $\frac{x+2}{x^{2}+4}$ có giá trị là số nguyên thì:
$\frac{x+2}{x^{2}+4}$ ⇒ $\frac{(x+2)(x-2)}{x^{2}-4}$ ⇒ $\frac{x^{2}+4-8}{x^{2}+4}$ ⇒ $\frac{8}{x^{2}+4}$
Xét $x^{2}$ + 4 = 4 ⇒ $x^{}$ = 0 (không thỏa).
Xét $x^{2}$ + 4 = 8 ⇔ $x^{}$ = ±2, thử lại ta thấy $x^{}$ = –2 thỏa yêu cầu bài toán.
Vậy $x^{}$ ∈ {–2}
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Đáp án+Giải thích các bước giải:
`\frac{x+2}{x^2+4}`
Để phân thức nguyên thì `x+2\vdots x^2+4`
`->x+2\vdots x^2+4`
`<=>(x-2)(x+2)\vdots x^2+4`
`<=>x^2-4\vdots x^2+4`
`<=>x^2+4-8\vdots x^2+4`
`<=>-8\vdots x^2+4`
`->x^2+4\inƯ(8)`
`->x^2+4\in{+-1;+-2;+-4;+-8}`
Mà `x^2+4>=4`
`->x^2+4=4`
`<=>x^2`
`<=>x=0`
`x^2+4=8`
`<=>x^2=2`
`<=>x=+-2`
Thử lại ta chỉ thấy `-2` là thỏa mãn
Vậy `x=-2` thì phân thức nguyên
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