Điều kiện $-\sqrt 2 \le x \le \sqrt 2.$+ Tìm max :Theo BĐT Bunhia$y^2=(x+\sqrt{2-x^2})^2 \le (1+1)(x^2+2-x^2)=4\Rightarrow -2 \le y \le 2.$Suy ra $\max y = 2 \Leftrightarrow x = 1.$+ Tìm min : Dễ thấy$\begin{cases}x\ge -\sqrt 2 \\ \sqrt{2-x^2} \ge 0 \end{cases}\Rightarrow y=x+\sqrt{2-x^2}\ge -\sqrt 2$Suy ra $\min y = -\sqrt 2 \Leftrightarrow x = -\sqrt 2.$
Điều kiện $-\sqrt 2 \le x \le \sqrt 2.$+ Tìm max :Theo BĐT Bunhia$y^2=(x+\sqrt{2-x^2})^2 \le (1+1)(x^2+2-x^2)=4\Rightarrow -2 \le y \le 2.$Suy ra $\max y = 2 \Leftrightarrow x = \pm 1.$+ Tìm min : Dễ thấy$\begin{cases}x\ge -\sqrt 2 \\ \sqrt{2-x^2} \ge 0 \end{cases}\Rightarrow y=x+\sqrt{2-x^2}\ge -\sqrt 2$Suy ra $\min y = -\sqrt 2 \Leftrightarrow x = -\sqrt 2.$
Điều kiện $-\sqrt 2 \le x \le \sqrt 2.$+ Tìm max :Theo BĐT Bunhia$y^2=(x+\sqrt{2-x^2})^2 \le (1+1)(x^2+2-x^2)=4\Rightarrow -2 \le y \le 2.$Suy ra $\max y = 2 \Leftrightarrow x = 1.$+ Tìm min : Dễ thấy$\begin{cases}x\ge -\sqrt 2 \\ \sqrt{2-x^2} \ge 0 \end{cases}\Rightarrow y=x+\sqrt{2-x^2}\ge -\sqrt 2$Suy ra $\min y = -\sqrt 2 \Leftrightarrow x = -\sqrt 2.$