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.
