Đáp án `+` Giải thích các bước giải:

Bài `3:`

`a,` Thay `x=-1` vào `A,` ta có:

`A=(3.(-1)+11)/(-1-3)=(-3+11)/(-4)=(-8)/(-4)=2`

Vậy `A=2` khi `x=-1`

`b, B=(x-1)/(x+3)+6/(x-3)-(2x-18)/(9-x^2)`

      `=(x^2-4x+3+6x+18+2x-18)/((x-3)(x+3))`

      `=(x^2+4x+3)/((x-3)(x+3))`

      `=((x+1)(x+3))/((x-3)(x+3))`

      `=(x+1)/(x-3)`

Vậy `B=(x+1)/(x-3)`

Bài `4:`

`a, P=(x^2-6x+9)/(9-x^2)+(4x+8)/(x+3)`

      `=(x-3)^2/(3-x)(x+3)+(4x+8)/(x+3)`

      `=-(x-3)(x+3)+(4x+8)/(x+3)`

      `=(4x+8-x+3)/(x+3)`

      `=(3x+11)/(x+3)`

Vậy `P=(3x+11)/(x+3)`

`b,` Thay `x=7` vào `P,` ta có:

`P=(3.7+11)/(7+3)=32/10=16/5`

Vậy `P=16/5` khi `x=7`

`c, P=(3x+11)/(x+3)=(3x+9+2)/(x+3)=(3x+9)/(x+3)+2/(x+3)=3+2/(x+3)`

Để `P` nguyên thì `2/(x+3)∈Z` hay `2\vdots(x+3)` hay `(x+3)∈Ư(2)={+-1;+-2}`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{x+3}&\text{-2}&\text{-1}&\text{1}&\text{2}\\\hline \text{x}&\text{-5(tm)}&\text{-4(tm)}&\text{-2(tm)}&\text{-1(tm)}\\\hline\end{array}

Vậy `x∈{-5;-4;-2;-1}` thì `P` có giá trị nguyên

`color{#66CCFF}{~Z}color{#66CCCC}{o}color{#66CC99}{n}color{#66CC66}{z}color{#66CC33}{o}color{#66CC00}{n}color{#33CCFF}{1}color{#33CCCC}{2}color{#33CC99}{3~}`