XD

`1.` Ta có:

`x^3 - 6x^2 + 12x - 11 = 0`

`⇔ x^3 - 3.2.x^2 + 3.2^2.x - 8 - 3 = 0`

`⇔ (x - 2)^3 - (\root{3}{3})^3 = 0`

`⇔ (x - 2 - \root{3}{3})[(x - 2)^2 + \root{3}{3}(x - 2) + (\root{3}{3})^2] = 0`

Vì: `(x - 2)^2 + \root{3}{3}(x - 2) + (\root{3}{3})^2\ > 0\ ∀ x`

`⇒ x - 2 - \root{3}{3} = 0`

`⇔ x = 2 + \root{3}{3}`

Vậy nghiệm của PT là: `S = {2 + \root{3}{3}}`

$\\$

`2.` Ta có:

`5x^2 - y^2 + 4xy - 9 = 0`

`⇔ 5x^2 + 4xy - y^2 = 9`

`⇔ 5x^2 + 5xy - xy - y^2 = 9`

`⇔ 5x(x + y) - y(x + y) = 9`

`⇔ (x + y)(5x - y) = 9 = 9.1 = (-9)(-1) = 3.3 = (-3)(-3)`

Ta có bảng sau:

$\begin{array}{|c|c|c|}\hline \text{5x - y}&\text{- 9}&\text{ 9}&\text{-3}&\text{3}&\text{-1}&\text{1}\\\hline \text{x + y}&\text{- 1}&\text{1}&\text{- 3}&\text{3}&\text{- 9}&\text{9}\\\hline \text{x}&\dfrac{-5}{3}&\dfrac{5}{3}&\text{- 1}&\text{1}&\dfrac{- 5}{3}&\dfrac{5}{3}\\\hline \text{y}&\dfrac{2}{3}&\dfrac{-2}{3}&\text{- 2}&\text{2}&\dfrac{- 22}{3}&\dfrac{22}{3}\\\hline\end{array}$

Mà: `x; y` là các số nguyên

`⇒ (x; y) = {(- 1; -2); (1; 2)}`

Vậy: `(x; y) = {(- 1; -2); (1; 2)}`