Đáp án + Giải thích các bước giải:
`a//` `ĐKXĐ:x\ne{-1;2;3}`
`M=(1-(x)/(1+x)):((x+3)/(x-2)+(x+2)/(3-x)+(x+2)/(x^{2}-5x+6))`
`=>M=((x+1)/(x+1)-(x)/(x+1)):(((x+3)(x-3))/((x-2)(x-3))-((x+2)(x-2))/((x-3)(x-2))+(x+2)/((x-2)(x-3)))`
`=>M=(1)/(x+1):(((x+3)(x-3)-(x+2)(x-2)+(x+2))/((x-2)(x-3)))`
`=>M=(1)/(x+1):((x^{2}-9-(x^{2}-4)+x+2)/((x-2)(x-3)))`
`=>M=(1)/(x+1):((x^{2}-9-x^{2}+4+x+2)/((x-2)(x-3)))`
`=>M=(1)/(x+1):(x-3)/((x-2)(x-3))`
`=>M=(1)/(x+1):(1)/(x-2)`
`=>M=(1)/(x+1).(x-2)/(1)`
`=>M=(x-2)/(x+1)`
`b//`
`M<0`
`<=>(x-2)/(x+1)<0`
`+)TH1:`
$\left\{\begin{matrix}x-2>0& \\x+1<0& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}x>2& \\x< -1& \end{matrix}\right.$
`=>2<x< -1` ( Loại )
`+)TH2:`
$\left\{\begin{matrix}x-2<0& \\x+1>0& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}x<2& \\x> -1& \end{matrix}\right.$
`=>-1<x<2` ( Nhận )
Vậy với `-1<x<2` thì `M<0`
`c//`
`M=(x-2)/(x+1)∈ZZ`
`=>x-2\vdots x+1`
`=>(x+1)-3\vdots x+1`
Vì `(x+1)\vdots x+1`
`=>3\vdots x+1`
`=>x+1∈Ư(3)={±1;±3}`
`=>x∈{0;2;-2;-4}`
Mà `ĐKXĐ:x\ne{-1;2;3}`
`=>x∈{0;-2;-4}`
