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popstar1102


A=$\sqrt{2-2x-x^2}$

ta có $A_{max}$ \Leftrightarrow $A^2_{max}$

\Rightarrow $A^2= 2-2x-x^2$

\Leftrightarrow $A^2=-(x^2+2x-2)=-(x^2-2x+1-3)$

\Leftrightarrow $A^2=-(x-1)^2+3$\leq 3

\Rightarrow $A^2_{max}$=3

\Rightarrow $A_{max}=\sqrt{3}$ \Leftrightarrow x=1
 
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