Toán học 9

L

leminhnghia1

Giải

Ta có:

[TEX]A=\frac{xy}{x+y+2}[/TEX]

[TEX] \Rightarrow \frac{1}{A}=\frac{xy}{x+y+2}=\frac{1}{x}+\frac{1}{y}+\frac{2}{xy} \ \geq \ 2\ sqrt{\frac{1}{xy}} \ + \ \frac{2}{xy}[/TEX]

Mà [TEX]xy \ \leq \ \frac{x^2+y^2}{2}=2[/TEX]

[TEX]\Rightarrow \ \frac{1}{A} \ \geq \ \frac{2}{ \ sqrt {2}}+\frac{1}{2}=\frac{1+2\ sqrt {2}}{2}[/TEX]

[TEX]\Rightarrow \ A \ \leq \ \frac{2}{2\ sqrt {2} \ + \ 1}[/TEX]

Vậy Max [TEX]A=\frac{2}{2\ sqrt {2} \ + \ 1} \ \Leftrightarrow \ x=y=\ sqrt {2}[/TEX]
 
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