Ai giúp với cần gấp ạ

Bài `7`.

`a) P = (x^2 - 6x + 9)/(9-x^2) + (8+4x)/(x+3) (x ne pm3)`

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

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

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

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

`=(3x+11)/(x+3)`

`b)` Thay `x=-2` (tmđk) vào `P`:

`P = (3.(-2)+11)/(3-2)`

`P = 5`

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

`-> 2/(x+3) in ZZ` để `P in ZZ`

`->` `x+3 in {pm1;pm2}`

`->` `x in {-5; -4; -2;-1}`

Đối chiếu đkxđ `-> x in {-5; -4; -2;-1}`

$\\$

Bài `8`.

`a) A = ( (2+x)/(2-x) - 4x^2/(x^2-4) - (2-x)/(2+x) ) : (x^2-3x)/(2x^2-x^3)`

`(x ne pm2; x ne 0; x ne 3)`

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

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

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

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

`= 4x/2-x . x(2-x)/(x-3)`

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

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

 `b) A = (4x^2)/(x-3)`

`A>0` `-> {(4x^2>0),(x-3>0):}` hoặc `{(4x^2<0),(x-3<0):}` (loại do `4x^2 >=0`)

`-> {(4x^2>0),(x-3>0):}`

`{(x ne 0),(x>3):}`

`-> x>3; x ne 0`

Vậy `A>0` khi `x>3`

`c) |x-7| = 4`

TH1: `x >= 7 -> x - 7 = 4`

`x = 11`

`-> A = (4.11^2)/(11-3) = 484/8 = 121/2`

TH2: `x < 7 -> 7-x = 4`

`x = 3` (không t/m đkxđ)

Vậy `A = 121/2` khi `|x-7|=4`