Cho hai số thực x, y thỏa mãn \(x + y + 1 = 2\left...

C

- Hướng dẫn giải

Điều kiện: \(\left\{ \begin{array}{l} x \ge 2\\ y \ge - 3 \end{array} \right.\), suy ra \(x + y + 1 \ge 0\).

● Ta có 

\(\begin{array}{l} x + y + 1 = 2\left( {\sqrt {x - 2} + \sqrt {y + 3} } \right)\,\\ = 2\sqrt {x - 2} + 2\sqrt {y + 3} \le \frac{{4 + x - 2}}{2} + \frac{{4 + y + 3}}{2} = \frac{{x + y + 9}}{2} \end{array}\)

Suy ra ​\(x + y + 1 \le \frac{{x + y + 9}}{2} \Leftrightarrow x + y \le 7\).

● Lại có \(x + y + 1 = 2\left( {\sqrt {x - 2} + \sqrt {y + 3} } \right)\)

\( \Leftrightarrow {\left( {x + y + 1} \right)^2} = 4\left( {x + y + 1 + 2\sqrt {x - 2} \sqrt {y + 3} } \right) \ge 4\left( {x + y + 1} \right)\) (do \(2\sqrt {x - 2} \sqrt {y + 3} \ge 0\))

Suy ra 

\({\left( {x + y + 1} \right)^2} \ge 4\left( {x + y + 1} \right) \\\Leftrightarrow \left[ \begin{array}{l} x + y + 1 \le 0\\ x + y + 1 \ge 4 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x + y + 1 = 0\\ x + y + 1 \ge 4 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x + y = - 1\\ x + y \ge 3 \end{array} \right..\)