giải hệ pt

B

braga

$$hpt\iff \begin{cases}1+\dfrac{1}{x+y}=\dfrac{3}{2\sqrt{x}} \ \ \ (1)\\1-\dfrac{1}{x+y}=\dfrac{1}{2\sqrt{y}} \ \ \ (2)\end{cases}$$
$(1)+(2)\implies \dfrac{3}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}=2 \ \ \ (\star) \\ (1)-(2)\implies \dfrac{3}{2\sqrt{x}}-\dfrac{1}{2\sqrt{y}}=\dfrac{1}{x+y} \ \ (\star \star) \\ (\star).(\star \star)\implies \dfrac{9}{4x}-\dfrac{1}{4y}=\dfrac{2}{x+y}$
 
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