Đáp án:
`a,Q=((2x-x^2)/(2x^2+8)-(2x^2)/(x^3-2x^2+4x-8))(2/(x^2)+(1-x)/x)`
(ĐKXĐ: `x\ne0;x\ne2`)
`\toQ=((2x-x^2)/(2(x^2+4))-(2x^2)/((x-2)(x^2+4)))(2/(x^2)+(x(1-x))/(x^2))`
`\toQ=(((x-2)(2x-x^2)-2.2x^2)/(2(x-2)(x^2+4)))((2+x-x^2)/(x^2))`
`\toQ=((-x^3-4x)/(2(x-2)(x^2+4)))((-x^2+x+2)/(x^2))`
`\toQ=(((-x)(x^2+4))/(2(x-2)(x^2+4)))((-x-1)(x-2))/(x^2)`
`\toQ=((-x)/(2(x-2)))((-x-1)(x-2))/(x^2)`
`\toQ=((-x)(-x-1)(x-2))/(2(x-2).x^2)`
`\toQ=(-x(-x-1))/(2x^2)`
`\toQ=(x^2+x)/(2x^2)`
`\toQ=(x(x+1))/(2x^2)`
`\toQ=(x+1)/(2x)`
`b,` Để `Q` nguyên
`\toQ\inZZ`
`\tox+1\vdots2x`
`\to2(x+1)\vdots2x`
`\to2x+2\vdots2x`
`\to2\vdots2x`
`\to2x\in Ư(2)={+-1;+-2}`
Ta có bảng:
\begin{array}{|c|c|}\hline 2x&1&-1&2&-2\\\hline x&\dfrac{1}{2}(\text{tm})&-\dfrac{1}{2}(\text{tm})&1(\text{tm})&-1(\text{tm})\quad\\\hline \end{array}
Vậy `x\in{1/2;-1/2;1;-1}`
