Tìm tất cả các cặp số `(x;y)` thỏa mãn sao cho `:` `5x-` 2 √x `(y+2)+y^2+1=0`

Đáp án:

$\left\{\begin{array}{l} y=\dfrac{1}{2} \\ x=\dfrac{1}{4}\end{array} \right..$

Giải thích các bước giải:

$5x-2\sqrt{x} (y+2)+y^2+1=0(x \ge 0)\\ \Leftrightarrow 5x-2\sqrt{x}y-4\sqrt{x}+y^2+1=0\\ \Leftrightarrow (x-2\sqrt{x}y+y^2)+(4x-4\sqrt{x}+1)=0\\ \Leftrightarrow (\sqrt{x}-y)^2+(2\sqrt{x}-1)^2=0\\ \text{Do } (\sqrt{x}-y)^2 \ge 0 \ \forall \ x \ge 0, y \in \mathbb{R}\\ (2\sqrt{x}-1)^2 \ge 0 \ \forall \ x \ge 0\\ \Rightarrow (\sqrt{x}-y)^2+(2\sqrt{x}-1)^2 \ge 0 \ \forall \ x \ge 0, y \in \mathbb{R}$

Dấu "=" xảy ra

$\Leftrightarrow \left\{\begin{array}{l} \sqrt{x}-y=0 \\ 2\sqrt{x}-1=0\end{array} \right.\Leftrightarrow \left\{\begin{array}{l} y=\sqrt{x}\\ 2\sqrt{x}=1\end{array} \right.\Leftrightarrow \left\{\begin{array}{l} y=\dfrac{1}{2} \\ \sqrt{x}=\dfrac{1}{2}\end{array} \right.\Leftrightarrow \left\{\begin{array}{l} y=\dfrac{1}{2} \\ x=\dfrac{1}{4}\end{array} \right..$