Question

Cho hệ phương trình sau:\(\left\{{}\begin{matrix}x+2\sqrt{xy}-2\sqrt{y}=1\\x+y=2019\end{matrix}\right.\)

Answer 1

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\y\ge0\end{matrix}\right.\)

- Từ PT ( I ) ta có : \(x-1+2\sqrt{xy}-2\sqrt{y}=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{y}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+1+2\sqrt{y}\right)=0\)

Thấy : \(\sqrt{x}+2\sqrt{y}+1\ge1>0\)

\(\Rightarrow\sqrt{x}-1=0\)

\(\Leftrightarrow x=1\)

- Thay x = 1 vào PT ( II ) ta được :

\(y=2019-x=2019-1=2018\)

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