Bài 5
[tex](3-x)(3-y)=9-3(x+y)+xy[/tex]
Đặt x+y=a
xy=b
Thep giả thiết ta có [tex]x^2+y^2=1\rightarrow a^2-2b=1\rightarrow b=\frac{a^2-1}{2}[/tex]
suy ra
[tex]9-3(x+y)+xy=9-3a+b=9+\frac{a^2-1}{2}-3a=\frac{a^2}{2}+\frac{17}{2}-3a=\frac{1}{2}(a^2+17)-3a=\frac{1}{2}(a^2-2\sqrt{2}a+2+15+2\sqrt{2}a)-3a=\frac{1}{2}(a-\sqrt{2})^2+\frac{15}{2}+\sqrt{2}a-3a=\frac{1}{2}(a-\sqrt{2})^2+\frac{15}{2}+(\sqrt{2}-3)a[/tex]
Có:
[tex]1=x^2+y^2\geq \frac{(x+y)^2}{2}=\frac{a^2}{2}\rightarrow a^2\leq 2\rightarrow a\leq 2[/tex]
Vậy [tex]\frac{1}{2}(a-\sqrt{2})^2+\frac{15}{2}+(\sqrt{2}-3)a\geq \frac{1}{2}(a-\sqrt{2})^2+\frac{15}{2}+(\sqrt{2}-3).\sqrt{2}[/tex] (Do [tex]\sqrt{2}-3<0[/tex])[tex]\frac{1}{2}(a-\sqrt{2})^2+\frac{15}{2}+(\sqrt{2}-3).\sqrt{2}\geq \frac{1}{2}(a-\sqrt{2})^2+\frac{19-6\sqrt{2}}{2}\geq \frac{19-6\sqrt{2}}{2}[/tex]
[tex]"="\rightarrow x=y=\frac{1}{\sqrt{2}}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
