Tìm `P_{max}` `P = ( \sqrt{x} - 1 )/( x - 2\sqrt{x} + 2 )`

`P = (sqrtx - 1)/(x - 2sqrtx + 2)` ĐK `: x >= 0`

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

`= (x - 2sqrtx + 2 - x + 4sqrtx - 4)/(2(x - 2sqrtx + 2))`

`= 1/2 + (-x + 4sqrtx - 4)/(2(x - 2sqrtx + 2))`

`= 1/2 - (x - 4sqrtx + 4)/(2(x - 2sqrtx + 2))`

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

Ta có `: (sqrtx - 2)^2 >= 0`

Mà `x - 2sqrtx + 2 = (sqrtx - 1)^2 + 1 >=1 > 0`

`=> (sqrtx - 2)^2/(2(x - 2sqrtx + 2)) >= 0`

`=> - (sqrtx - 2)^2/(2(x - 2sqrtx + 2)) <= 0`

`=>1/2 - (sqrtx - 2)^2/(2(x - 2sqrtx + 2))  <= 1/4`

Dấu "=" xảy ra khi `: sqrtx - 2 = 0 <=> x = 4(tm)`

Vậy `P_(max) = 1/2 <=> x = 4`