Question

Answer 1

`a,` Với ĐKXĐ: `x \ne +-3,` ta có:

`P = (2x)/(x + 3)+(x + 1)/(x -3) + (3-11x)/(9-x^2)`

`= (2x)/(x + 3) + (x + 1)/(x - 3) + (3 - 11x)/((3 - x)(3 + x ))`

`= (2x)/(x + 3) + (x + 1)/(x - 3)  - (3 - 11x)/((x + 3)(x - 3))`

`= (2x . (x - 3))/((x + 3)(x - 3)) + ((x + 1)(x  + 3))/((x + 3)(x - 3)) - (3  -11x)/((x + 3)(x - 3))`

`= (2x . (x - 3) + (x + 1)(x + 3) - (3  -11x))/((x + 3)(x - 3)`

`= (2x^2 - 6x + x^2 + 4x + 3 - 3 + 11x)/((x + 3)(x - 3))`

`= (3x^2 + 9x)/((x + 3)(x - 3))`

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

`= (3x)/(x - 3) (đpcm)`

 

Answer 2

`b,` Ta có: `P = (3x)/(x - 3) = (3x - 9 + 9)/(x - 3) = 3 + 9/(x-  3)`

Để `x` nguyên thì `9 \vdots (x - 3),` ta có:

`9 \vdots (x - 3)` hay `x - 3 \in Ư(9)={+-1;+-3;+-9}`

Vì để `x` nhận giá trị nguyên nên ta nhận `x - 3 = {1;3;9}`

Suy ra: `x = {4;6;12} (TM)`

Vậy: `x ={4;6;12}` thì `P` có giá trị nguyên dương.

Answer 3

`c,` Ta có: `Q = P . (x- 3)/5.` hay `(3x)/(x - 3) . (x - 3)/5 = 1/5`

`(3x)/5 = 1/5`

`3x = 1`

`x = 1 : 3`

`x = 1/3`

Vậy: `x = 1/3`