`a)` Với `x \ne 1` có:
`C=[5x+1]/[x^3-1]-[1-2x]/[x^2+x+1]-2/[1-x]`
`C=[5x+1-(1-2x)(x-1)+2(x^2+x+1)]/[(x-1)(x^2+x+1)]`
`C=[5x+1-x+1+2x^2-2x+2x^2+2x+2]/[(x-1)(x^2+x+1)]`
`C=[4x^2+4x+4]/[(x-1)(x^2+x+1)]`
`C=[4(x^2+x+1)]/[(x-1)(x^2+x+1)]=4/[x-1]`
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`b)x^2-2x=0`
`<=>x(x-2)=0`
`<=>` $\left[\begin{matrix} x=0\\ x=2\end{matrix}\right.$ (t/m)
`@` Thay `x=0` vào `C` có: `C=4/[0-1]=-4`
`@` Thay `x=2` vào `C` có: `C=4/[2-1]=4`
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`c)` Với `x \ne 1` có:
`C > 0<=>4/[x-1] > 0` Mà `4 > 0`
`=>x-1 > 0<=>x > 1` (t/m `x \ne 1`)
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`d)` Với `x \ne 1` có: `C=4/[x-1]`
Để `C in ZZ<=>4/[x-1] in ZZ`
`=>x-1 in Ư_4`
Mà `Ư_4={+-1;+-2;+-4}`
Ta có bảng:
\begin{array}{|c|c|c|}\hline x-1&-1&1&-2&2&-4&4\\\hline x&0&2&-1&3&-3&5 \\\hline\end{array}
Mà `x in ZZ,x \ne 1=>x in {0;2;-1;3;-3;5}`