Đáp án:
$\left \{ {\dfrac{1}{x + 1} -2\sqrt{y} = -1{} \atop {\dfrac{2(\sqrt{y} + 1)}{x + 1} + y - 2\sqrt{y} - 3 = 0}} \right.$ `↔` $\left \{ {\dfrac{1}{x + 1} = 2\sqrt{y} -1{} \atop {\dfrac{2(\sqrt{y} + 1)}{x + 1} + y - 2\sqrt{y} - 3 = 0}} \right.$
`↔` $\left \{ {\dfrac{1}{x + 1} = 2\sqrt{y} -1{} \atop {2(\sqrt{y} + 1)(2\sqrt{y} - 1) + y - 2\sqrt{y} - 3 = 0}} \right.$
`↔` $\left \{ {\dfrac{1}{x + 1} = 2\sqrt{y} -1{} \atop {5y - 5 = = 0}} \right.$ `↔` $\left \{ {\dfrac{1}{x + 1} = 2\sqrt{y} -1{} \atop {y = 1}} \right.$ `↔` $\left \{ {x = 0{} \atop {y = 1}} \right.$
Vậy `x = 0 ; y = 1`
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