Phần c câu 5 ạ Mong các cao nhân giải nhanh giúp e

Câu `5`:

`a)`

`P=[3x+5sqrtx-11]/[(sqrtx-1)(sqrtx+2)]-[sqrtx-2]/[sqrtx-1]+2/[sqrtx+2]-1 (0<=x ne 1)`

`P=[3x+5sqrtx-11]/[(sqrtx-1)(sqrtx+2)]-[(sqrtx-2)(sqrtx+2)]/[(sqrtx-1)(sqrtx+2)]+[2(sqrtx-1)]/[(sqrtx-1)(sqrtx+2)]-[(sqrtx-1)(sqrtx+2)]/[(sqrtx-1)(sqrtx+2)]`

`P=[3x+5sqrtx-11-x+4+2sqrtx-2-x-2sqrtx+sqrtx+2]/[(sqrtx-1)(sqrtx+2)]`

`P=[x+6sqrtx-7]/[(sqrtx-1)(sqrtx+2)]`

`P=[(sqrtx-1)(sqrtx+7)]/[(sqrtx-1)(sqrtx+2)]`

`P=[sqrtx+7]/[sqrtx+2]`

`b)`

Thay `x=21+8sqrt5` vào `P` ta có:

`P=[sqrt[21+8sqrt5]+7]/[sqrt[21+8sqrt5]+2]`

`P=[sqrt[16+2*4sqrt5+5]+7]/[sqrt[16+2*4sqrt5+5]+2]`

`P=[sqrt[(4+sqrt5)^2]+7]/[sqrt[(4+sqrt5)^2]+2]`

`P=[4+sqrt5+7]/[4+sqrt5+2]`

`P=[11+sqrt5]/[6+sqrt5]`

`P=[(11+sqrt5)(6-sqrt5)]/[36-5]`

`P=[66-11sqrt5+6sqrt5-5]/31`

`P=[61-5sqrt5]/31`

Vậy ...

`c)`

`P vdots 3` nên

`[sqrtx+7]/[sqrtx+2] vdots 3`

`=> [sqrtx+7]/[sqrtx+2] in Ư(3)`

`=> [sqrtx+7]/[sqrtx+2] in {+-1;+-3}`

Giải các trường hợp trên, ta có `x=1/4(text[TM])`.

Vậy `x=1/4` thỏa mãn ycđb.

#BKT