`\color{#fc1100}{♡}\color{#ff483b}{T}\color{#ff655a}{u}\color{#ff857c}{ấ}\color{#feb8b3}{n}`

Bài `4.`

`bb(1.)`

Thay `x = 49 (tmđk)` vào `B :`

` B = (sqrt{49} + 1)/(sqrt{49} - 1) = (7 + 1)/(7 - 1) = 8/6 = 4/3`

`bb(2.)`
`A = (x - 2)/(sqrtx . (sqrtx + 2)) + 1/(sqrtx + 2)`

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

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

`A = (sqrtx - 1)/sqrtx`

`P = A . B:`

`to P = (sqrtx - 1)/sqrtx . (sqrtx + 1)/(sqrtx - 1) = (sqrtx + 1)/sqrtx`

`bb(c.)`

Ta có `:`

`P = (sqrtx + 1)/sqrtx`

Vì `x > 0 `

`to sqrtx > 0` và `sqrtx + 1 > 0`

`to P > 0`

`to 2P > 0`

`to |2P| = 2P`

` |2P| = 2 . sqrtx + 5:to 2 . ((sqrtx + 1)/sqrtx) = 2 . sqrtx + 5`

`to (2 . sqrtx + 2)/sqrtx = 2 . sqrtx + 5`

`to 2 . sqrtx + 2 = sqrtx . (2 . sqrtx + 5)`

`to 2 . sqrtx + 2 = 2x + 5 . sqrtx`

`to 2x + 3 . sqrtx - 2 = 0`

`to (2 . sqrtx - 1)(sqrtx + 2) = 0`

Vì `sqrtx + 2 > 0`

`to 2 . sqrtx - 1 = 0`

`to 2 . sqrtx = 1`

`to sqrtx = 1/2`

`to x = 1/4 (tmđk)`

Vậy `x = 1/4 .`