Tìm các giá trị nguyên của x để A nhân giá trị nguyên.

Giúp em với em đang cần gắp!! :-(

 `ĐKXĐ:`  `x≥0, x≠1`

`A=(\sqrt{x}+1)/(\sqrt{x}-1)+(\sqrt{x}-1)/(\sqrt{x}+1)-(3\sqrt{x}+1)/(x-1)`

   `=( (\sqrt{x}+1)^2 + (\sqrt{x}-1)^2 - (3\sqrt{x}+1) )/( (\sqrt{x}-1)(\sqrt{x}+1) )`

   `= (x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1)/( (\sqrt{x}-1)(\sqrt{x}+1) )`

   `=(2x-3\sqrt{x}+1)/( (\sqrt{x}-1)(\sqrt{x}+1) )`

   `=(2x-2\sqrt{x}-\sqrt{x}+1)/( (\sqrt{x}-1)(\sqrt{x}+1) )`

   `=( 2\sqrt{x}(\sqrt{x}-1)-(\sqrt{x}-1) )/( (\sqrt{x}-1)(\sqrt{x}+1) )`

   `= ( (\sqrt{x}-1)(2\sqrt{x}-1) )/( (\sqrt{x}-1)(\sqrt{x}+1) )`

   `= (2\sqrt{x}-1)/(\sqrt{x}+1)`

   `= ( 2(\sqrt{x}+1) -3 )/(\sqrt{x}+1)`

   `= 2- 3/(\sqrt{x}+1)`

`→` `3/(\sqrt{x}+1)∈ZZ` thì `A∈ZZ`

`⇒ (\sqrt{x}+1)∈Ư(3)={±1;±3}`

`⇒ (\sqrt{x}+1)∈{1;3}`  `(` Vì `\sqrt{x}+1≥1` `)`

`⇒\sqrt{x}∈{0;2}`

`⇒ x∈{0;4}`  `(TMĐK)`

Vậy khi `x∈{0;4}` thì `A` nhận giá trị nguyên

`#` `ccUh`

Ta có:

`A = (sqrtx + 1)/(sqrtx - 1) + (sqrtx - 1)/(sqrtx + 1) - (3 sqrtx + 1)/(x-1)` `(x >= 0; x ne 1)`

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

`= ((sqrtx+1)^2 + (sqrtx-1)^2 - 3 sqrtx - 1)/((sqrtx - 1)(sqrtx + 1))`

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

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

`= (2x - 2 sqrtx - sqrtx + 1)/((sqrtx - 1)(sqrtx + 1))`

`= (2 sqrtx(sqrtx - 1) - (sqrtx - 1))/((sqrtx - 1)(sqrtx + 1))`

`= ((sqrtx -1 )(2 sqrtx - 1))/((sqrtx - 1)(sqrtx + 1))`

`= (2 sqrtx - 1)/(sqrtx + 1)`

`= (2 sqrtx + 2 - 3)/(sqrtx + 1)`

`= (2(sqrtx + 1) - 3)/(sqrtx + 1)`

`= (2(sqrtx + 1))/(sqrtx+1) - 3/(sqrtx + 1)`

`= 2 - 3/(sqrtx + 1)`

Để `A` nguyên thì `3/(sqrtx + 1)` nguyên

`-> sqrtx + 1 in Ư(3) = {-3;-1;1;3}`

`-> sqrtx in {-4;-2;0;2}`

mà `sqrtx >= 0` nên `sqrtx in {0;2} -> x in {0;4}` (TMĐK)

Vậy `x in {0;4}` thì `A` nguyên

`\ttcolor{cyan}{#PUyen}`