Đáp án+Giải thích các bước giải:

`{( 3/\sqrt{y} - 1/x = (5x + \sqrt{y})/(2x^2 + y) (1) ),(1/(xy) + 4/\sqrt{y} = 2/y + 8/3 (2) ):}` (đkxđ: `y > 0 ; x \ne 0`)

Từ `(1) -> (3x - \sqrt{y})/(x\sqrt{y}) = (5x + \sqrt{y})/(2x^2 + y)`

`-> (3x - \sqrt{y})(2x^2 + y) = (5x + \sqrt{y})x\sqrt{y}`

`-> 6x^3 + 3xy - 2x^2\sqrt{y} - y\sqrt{y} - 5x^2\sqrt{y} - xy = 0`

`-> 6x^3 - 7x^2\sqrt{y} + 2xy - y\sqrt{y} = 0`

`-> 6x^2(x - \sqrt{y}) - x\sqrt{y}(x - \sqrt{y}) + y(x - \sqrt{y}) = 0`

`-> (x - \sqrt{y})(6x^2 - x\sqrt{y} + y) = 0`

`-> (x - \sqrt{y})[5x^2 + (x - 1/2\sqrt{y})^2 + 3/4y ] = 0`

`-> x - \sqrt{y} = 0` hoặc `5x^2 + (x - 1/2\sqrt{y})^2 + 3/4 = 0` (Vô lý)

`-> x = \sqrt{y}` (đk: `x > 0`)

`-> x^2 = y`

Thay `x^2 = y` vào `(2)` ta đc:

`1/x^3 + 4/x = 2/x^2 + 8/3`

`(3 + 12x^2)/(3x^3) = (6x + 8x^3)/(3x^3)`

`3 + 12x^2 = 6x + 8x^3`

`8x^3 - 12x^2 + 6x - 3 = 0`

`(8x^3 - 12x^2 + 6x - 1) - 2 = 0`

`(2x - 1)^3 = 2`

`2x - 1 = \root{3}{2}`

`2x = \root{3}{2} + 1`

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

`-> y = x^2 = ( (\root{3}{2} + 1)/2)^2`

Vậy ...