Bạn kia làm đc câu 1 r thì mình làm câu 2 nhé, câu 1 trình bày hơi dài
$HPT ⇔ \left \{ {{2\sqrt[]{xy} - \frac{\sqrt[]{xy} }{2x + y} = 2\sqrt[]{x} (1) } \atop {2\sqrt[]{xy} - \frac{\sqrt[]{xy} }{2x + y} = 2\sqrt[]{y} (2)}} \right. ⇔ \left \{ {{ (1) + (2) : \sqrt[]{y} - \sqrt[]{x} = 2\sqrt[]{xy} (3) } \atop { (2) - (1): \sqrt[]{y} + \sqrt[]{x} = \frac{\sqrt[]{xy} }{2x + y} (4)}} \right. ⇔ \left \{ {{ \sqrt[]{y} - \sqrt[]{x} = 2\sqrt[]{xy}} \atop { (3).(4) : y - x = \frac{2xy}{2x + y}}} \right.$
$⇔ \left \{ {{ \sqrt[]{y} - \sqrt[]{x} = 2\sqrt[]{xy}} \atop {xy - 2x² + y² = 2xy }} \right. ⇔ \left \{ {{ \sqrt[]{y} - \sqrt[]{x} = 2\sqrt[]{xy}} \atop {(x + y)(2x - y) = 0 }} \right. ⇔ \left \{ {{ \sqrt[]{y} - \sqrt[]{x} = 2\sqrt[]{xy}} \atop {y = 2x}} \right.$
$⇔ \left \{ {{ \sqrt[]{2x} - \sqrt[]{x} = 2\sqrt[]{2x²}} \atop {y = 2x}} \right. ⇔ \left \{ {{ x = (\frac{2\sqrt[]{2}}{\sqrt[]{2} - 1})²} \atop {y = 2x}} \right.⇔ \left \{ {{ x = 8(3 + 2\sqrt[]{2})}\atop {y = 16(3 + 2\sqrt[]{2})}} \right.$
