`#Ju ly `

`2. `

`a) ` Thay `x=4(TMDK) ` vào `A, ` ta có`: `

`A=(4+2)/(4+1)=6/5 `

Vậy ...

`b)P=A+B=(x+2)/(x+1)+[3/(x+3)-(6x+8)/(x^2+4x+3)] `

`=(x+2)/(x+1)+[3/(x+3)-(6x+8)/(x^2+x+3x+3)] `

`=(x+2)/(x+1)+[3/(x+3)-(6x+8)/(x(x+1)+3(x+1))] `

`=(x+2)/(x+1)+[3/(x+3)-(6x+8)/((x+1)(x+3))] `

`=(x+2)/(x+1)+(3(x+1)-6x-8)/((x+1)(x+3)) `

`=(x+2)/(x+1)+(3x+3-6x-8)/((x+1)(x+3)) `

`=(x+2)/(x+1)+(-3x-5)/((x+1)(x+3)) `

`=((x+2)(x+3)-3x-5)/((x+1)(x+3)) `

`=(x^2+5x+6-3x-5)/((x+1)(x+3)) `

`=(x^2+2x+1)/((x+1)(x+3)) `

`=((x+1)^2)/((x+1)(x+3)) `

`=(x+1)/(x+3)(đpcm) `

`c) ` Để `P=-1 ` thì`: `

`(x+1)/(x+3)=-1 `

`=>-(x+1)=x+3 `

`=>-x-1-x-3=0 `

`=>-2x-4=0 `

`=>-2x=4 `

`=>x=-2(TM) `

Vậy ...

`d) ` Ta có`:P=(x+1)/(x+3)=1-2/(x+3) `

Để `P  in  ZZ ` thì `1-2/(x+3)  in  ZZ `

`=>(x+3)  in  Ư(-2)={+-1;+-2} `

`=>x  in  {-2;-4;-5;-1} `

Kết hợp `ĐKXĐ:x ` $\ne$ `-1;x ` $\ne$ `-3 `

`=>x  in  {-2;-4;-5} `

Vậy ...