Đáp án: Giải thích các bước giải: $A=x+y+\frac{9}{x+y}+\frac{1}{x+y}=x+y+\frac{9}{x+y}+\frac{1}{x+\frac{8-x^2}{4}}$ $A=x+y+\frac{9}{x+y}+\frac{4}{12-(x-2)^2}$ $12-(x-2)^2=4x+4y>0 \rightarrow \frac{4}{12-(x-2)^2} \geq \frac{4}{12}=\frac{1}{3}$ $\rightarrow A \geq 2\sqrt{\frac{9(x+y)}{x+y}}+\frac{1}{3}=\frac{19}{3}$ $A_{min}=\frac{19}{3}$ khi $x=2; y=1$ Reply
Đáp án:
Giải thích các bước giải:
$A=x+y+\frac{9}{x+y}+\frac{1}{x+y}=x+y+\frac{9}{x+y}+\frac{1}{x+\frac{8-x^2}{4}}$
$A=x+y+\frac{9}{x+y}+\frac{4}{12-(x-2)^2}$
$12-(x-2)^2=4x+4y>0 \rightarrow \frac{4}{12-(x-2)^2} \geq \frac{4}{12}=\frac{1}{3}$
$\rightarrow A \geq 2\sqrt{\frac{9(x+y)}{x+y}}+\frac{1}{3}=\frac{19}{3}$
$A_{min}=\frac{19}{3}$ khi $x=2; y=1$