`A= x/((\sqrt{x}+\sqrt{y})(1-\sqrt{y}))-y/((\sqrt{x}+\sqrt{y})(\sqrt{x}+1))-(xy)/((\sqrt{x}+1)(1-\sqrt{y}))` a, Rút gọn b, Tìm $x, y$ nguyên thỏa mãn

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1. Đáp án:

   a) $A = \sqrt[]{xy}+\sqrt[]{x}-\sqrt[]{y}$ với $x ≥0, y>0$

   b) $(x;y) \in \big\{(2;2),(0;4)\big\}$ 

    Giải thích các bước giải:

    $ĐKXĐ : x ≥ 0, y >0 $

   a) Rút gọn :

   $A = \dfrac{x}{(\sqrt[]{x}+\sqrt[]{y}).(1-\sqrt[]{y})} – \dfrac{y}{(\sqrt[]{x}+\sqrt[]{y}).(\sqrt[]{x}+1)} – \dfrac{xy}{(\sqrt[]{x}+1).(1-\sqrt[]{y})}$

   $ = \dfrac{x.(\sqrt[]{x}+1)- y.(1-\sqrt[]{y}) – xy.(\sqrt[]{x}+\sqrt[]{y})}{(\sqrt[]{x}+\sqrt[]{y}).(1-\sqrt[]{y}).(\sqrt[]{x}+1)}$

   Xét $x.(\sqrt[]{x}+1)- y.(1-\sqrt[]{y}) – xy.(\sqrt[]{x}+\sqrt[]{y})$

   $ = x\sqrt[]{x}+x-y+y\sqrt[]{y} – xy.(\sqrt[]{x}+\sqrt[]{y})$

   $ = x.(\sqrt[]{x}+\sqrt[]{y}) + y.(\sqrt[]{x}+\sqrt[]{y}) -xy.(\sqrt[]{x}+\sqrt[]{y}) – x\sqrt[]{y} – y\sqrt[]{x} + x-y$

   $ = (\sqrt[]{x}+\sqrt[]{y}).(x+y-xy) – \sqrt[]{xy}.(\sqrt[]{x}+\sqrt[]{y}) + (\sqrt[]{x}+\sqrt[]{y}).(\sqrt[]{x}-\sqrt[]{y})$

   $ = (\sqrt[]{x}+\sqrt[]{y}).(x+y-xy-\sqrt[]{xy} + \sqrt[]{x}-\sqrt[]{y})$

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

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

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

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

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

   Do đó :

   $A = \dfrac{(\sqrt[]{x}+\sqrt[]{y}).(\sqrt[]{x}+1).(1-\sqrt[]{y}).(\sqrt[]{xy}+\sqrt[]{x}-\sqrt[]{y})}{(\sqrt[]{x}+\sqrt[]{y}).(\sqrt[]{x}+1).(1-\sqrt[]{y})}$

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

   Vậy $A = \sqrt[]{xy}+\sqrt[]{x}-\sqrt[]{y}$ với $x ≥0, y>0$

   b) Ta có : $A = 2$ nên : $\sqrt[]{xy} + \sqrt[]{x} – \sqrt[]{y} = 2$

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

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

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

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

   Vì $\sqrt[]{x} ≥ 0$ nên $\dfrac{1}{\sqrt[]{y}-1} – 1 ≥ 0 $

   $⇔ \dfrac{2-\sqrt[]{y}}{\sqrt[]{y}-1} ≥ 0 $

   +) Trường hợp 1 : $\left\{ \begin{array}{l}2-\sqrt[]{y}≥0\\\sqrt[]{y}-1>0\end{array} \right.$ $\left\{ \begin{array}{l}\sqrt[]{y}≤2\\\sqrt[]{y}>1\end{array} \right.$ $1<\sqrt[]{y}≤2$

   $⇔ 1< y ≤ 4$

   +) Trường hợp 2 :  $\left\{ \begin{array}{l}2-\sqrt[]{y}≤0\\\sqrt[]{y}-1<0\end{array} \right.$ $\left\{ \begin{array}{l}\sqrt[]{y}≥2\\\sqrt[]{y}<1\end{array} \right.$ ( Vô lí )

   Do đó : $1<y≤4$. Mà $y$ nguyên nên $y \in \big\{2,3,4\big\}$

   $·$ Với $y=2$ thì ta có : $\sqrt[]{x} = \dfrac{1}{\sqrt[]{2}-1} – 1 = \sqrt[]{2}$

   $⇒x=2$ ( Thỏa mãn )

   $·$ Với $y=3$ thì ta có : $\sqrt[]{x} = \dfrac{1}{\sqrt[]{3}-1} – 1= \dfrac{-1+\sqrt[]{3}}{2}$

   $ ⇒ x= \dfrac{2-\sqrt[]{3}}{2}$ ( Loại do $x$ nguyên )

   $·$ Với $y=4$ thì ta có : $\sqrt[]{x} = \dfrac{1}{\sqrt[]{4}-1} – 1 = 0$

   $⇒x=0$ ( Thỏa mãn )

   Vậy $(x;y) \in \big\{(2;2),(0;4)\big\}$ thỏa mãn đề.

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