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max A=[tex]\sqrt{x-1}+ [tex]\sqrt{y-2} khi cho: x+y=4[/QUOTE] Áp dụng BDT Bunhiacopxki ta có: [TEX]A^2=(\sqrt{x-1}+\sqrt{y-2})^2 \leq (1^2+1^2)(x-1+y-2)=2[/TEX]
[TEX]\Rightarrow A \leq \sqrt{2}[/TEX]
[TEX]MaxA=\sqrt{2},khi:\left{\begin{\sqrt{x-1}=\sqrt{y-2}}\\{x+y=4} \Leftrightarrow \left{\begin{x=\frac{3}{2}}\\{y=\frac{5}{2}}[/TEX]
 
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