Đáp án + Giải thích các bước giải:

$a.$ ĐKXĐ :

+) $x ≥ 0 , y ≥ 0$

+) $( \sqrt[]{x} + \sqrt[]{y} )( 1 - \sqrt[]{y} ) \ne 0 $

+) $( \sqrt[]{x} + \sqrt[]{y} )( \sqrt[]{x} + 1 ) \ne 0$

Vì $\sqrt[]{x} + 1 \ne 0$ với $∀ x ≥ 0$

⇒ $\sqrt[]{x} + \sqrt[]{y} \ne 0$

⇒ $x \ne y \ne 0$

$+) ( \sqrt[]{x} + 1 )( 1 - \sqrt[]{y} ) \ne 0$

Vì $\sqrt[]{x} + 1 \ne 0$ với $∀ x ≥ 0$

⇒ $1 - \sqrt[]{y} \ne 0$

⇔ $\sqrt[]{y} \ne 1$

⇔ $y \ne 1$

Kết hợp các điều kiện xác định :

⇒ $x ≥ 0 , y ≥ 0 , x \ne y \ne 0 , y \ne 1$

$b.$

$P = \frac{x}{(\sqrt[]{x}+\sqrt[]{y})(1-\sqrt[]{y})} - \frac{y}{(\sqrt[]{x}+\sqrt[]{y})(\sqrt[]{x}+1)} - \frac{xy}{(\sqrt[]{x}+1)(1-\sqrt[]{y})}$

$P = \frac{x(\sqrt[]{x}+1)-y(1-\sqrt[]{y})-xy(\sqrt[]{x}+\sqrt[]{y})}{(\sqrt[]{x}+\sqrt[]{y})(1-\sqrt[]{y})(\sqrt[]{x}+1)}$

$P = \frac{x\sqrt[]{x}+x-y+y\sqrt[]{y}-xy(\sqrt[]{x}+\sqrt[]{y})}{(\sqrt[]{x}+\sqrt[]{y})(1-\sqrt[]{y})(\sqrt[]{x}+1)}$

$P = \frac{(\sqrt[]{x})^{3}+(\sqrt[]{y})^{3}+(x-y)-xy(\sqrt[]{x}+\sqrt[]{y})}{(\sqrt[]{x}+\sqrt[]{y})(1-\sqrt[]{y})(\sqrt[]{x}+1)}$

$P = \frac{(\sqrt[]{x}+\sqrt[]{y})(x-\sqrt[]{xy}+y)+(\sqrt[]{x}+\sqrt[]{y})(\sqrt[]{x}-\sqrt[]{y})-xy(\sqrt[]{x}+\sqrt[]{y})}{(\sqrt[]{x}+\sqrt[]{y})(1-\sqrt[]{y})(\sqrt[]{x}+1)}$

$P = \frac{(\sqrt[]{x}+\sqrt[]{y})(x-\sqrt[]{xy}+y+\sqrt[]{x}-\sqrt[]{y}-xy)}{(\sqrt[]{x}+\sqrt[]{y})(1-\sqrt[]{y})(\sqrt[]{x}+1)}$

$P = \frac{x-\sqrt[]{xy}+\sqrt[]{x}-\sqrt[]{y}-xy}{(1-\sqrt[]{y})(\sqrt[]{x}+1)}$

$P = \frac{(x+\sqrt[]{x})-(\sqrt[]{xy}+\sqrt[]{y})-(xy-y)}{(1-\sqrt[]{y})(\sqrt[]{x}+1)}$

$P = \frac{\sqrt[]{x}(\sqrt[]{x}+1)-\sqrt[]{y}(\sqrt[]{x}+1)-y(x-1)}{(1-\sqrt[]{y})(\sqrt[]{x}+1)}$

$P = \frac{(\sqrt[]{x}+1)(\sqrt[]{x}-\sqrt[]{y})-y(\sqrt[]{x}+1)(\sqrt[]{x}-1)}{(1-\sqrt[]{y})(\sqrt[]{x}+1)}$

$P = \frac{(\sqrt[]{x}+1)(\sqrt[]{x}-\sqrt[]{y}-\sqrt[]{x}y+y)}{(1-\sqrt[]{y})(\sqrt[]{x}+1)}$

$P = \frac{\sqrt[]{x}-\sqrt[]{y}-\sqrt[]{x}y+y}{1-\sqrt[]{y}}$

$P = \frac{(\sqrt[]{x}-\sqrt[]{x}y)-(\sqrt[]{y}-y)}{1-\sqrt[]{y}}$

$P = \frac{\sqrt[]{x}(1-y)-\sqrt[]{y}(1-\sqrt[]{y})}{1-\sqrt[]{y}}$

$P = \frac{\sqrt[]{x}(1-\sqrt[]{y})(1+\sqrt[]{y})-\sqrt[]{y}(1-\sqrt[]{y})}{1-\sqrt[]{y}}$

$P = \frac{(1-\sqrt[]{y})(\sqrt[]{x}+\sqrt[]{xy}-\sqrt[]{y})}{1-\sqrt[]{y}}$

$P = \sqrt[]{x} + \sqrt[]{xy} - \sqrt[]{y}$

Theo bài ta có : $P = 2$

⇔ $\sqrt[]{x} + \sqrt[]{xy} - \sqrt[]{y} = 2$

⇔ $\sqrt[]{x}( 1 + \sqrt[]{y} ) - ( \sqrt[]{y} + 1 ) = 1$

⇔ $( \sqrt[]{y} + 1 )( \sqrt[]{x} - 1 ) = 1$

⇔ $\sqrt[]{x} - 1 = \frac{1}{\sqrt[]{y}+1}$

Vì $\sqrt[]{y} + 1 ≥ 1$ với $∀ y ≥ 0$

⇒ $\frac{1}{\sqrt[]{y}+1} ≤ 1$

⇒ $\sqrt[]{x} - 1 ≤ 1$

Vì $\sqrt[]{x} - 1 ≥ - 1$ với $∀ x ≥ 0$

⇒ $- 1 ≤ \sqrt[]{x} - 1 ≤ 1$

⇔ $0 ≤ \sqrt[]{x} ≤ 2$

⇔ $0 ≤ x ≤ 4$

Vì $x ∈ Z$

⇒ $x =$ {$0 , 1 , 2 , 3 , 4$}
+) $x = 0 ⇒ - \sqrt[]{y} - 1 = 1 ⇔ \sqrt[]{y} = - 2$

⇒ Loại

+) $x = 1 ⇒ ( \sqrt[]{y} + 1 )( 1 - 1 ) = 1 ⇔ ( \sqrt[]{y} + 1 ).0 = 1$

⇒ Loại

+) $x = 2 ⇒ ( \sqrt[]{y} + 1 )( \sqrt[]{2} - 1 ) = 1$

⇔ $\sqrt[]{y} + 1 = \frac{1}{\sqrt[]{2}-1}$

⇔ $\sqrt[]{y} + 1 = 1 + \sqrt[]{2}$

⇔ $\sqrt[]{y} = \sqrt[]{2}$

⇔ $y = 2$

+) $x = 3 ⇒ ( \sqrt[]{y} + 1 )( \sqrt[]{3} - 1 ) = 1$

⇔ $\sqrt[]{y} + 1 = \frac{1}{\sqrt[]{3}-1}$

⇔ $\sqrt[]{y} + 1 = \frac{1+\sqrt[]{3}}{2}$

⇔ $\sqrt[]{y} = \frac{-1+\sqrt[]{3}}{2}$

⇔ $y = \frac{2-\sqrt[]{3}}{2}$

⇒ Loại vì $y ∈ Z$

+) $x = 4 ⇒ ( \sqrt[]{y} + 1 )( 2 - 1 ) = 1$

⇔ $\sqrt[]{y} + 1 = 1$

⇔ $\sqrt[]{y} = 0$

⇔ $y = 0$

Kết hợp các trường hợp ⇒ $( x , y ) = ( 2 , 2 ) ; ( 4 , 0 )$