Tìm min, max của P= [tex]\frac{\sqrt{1+x}}{\sqrt{x}+\sqrt{1-x}}[/tex]
*min: [tex]\frac{\sqrt{1+x}}{\sqrt{x}+\sqrt{1-x}}\\\\ =\frac{\sqrt{2.}\sqrt{1+x}}{\sqrt{2x}+\sqrt{2}\sqrt{1-x}}\\\\ +, (\sqrt{2x}+\sqrt{2}\sqrt{1-x})^2\leq (1+2).(2x+1-x)=3.(x+1)\\\\ => \frac{\sqrt{2.}\sqrt{1+x}}{\sqrt{2x}+\sqrt{2}\sqrt{1-x}}\geq \frac{\sqrt{2}.\sqrt{x+1}}{\sqrt{3}.\sqrt{x+1}}=\frac{\sqrt{2}}{\sqrt{3}}[/tex]
dấu "=" <=> [tex]\frac{\sqrt{2x}}{1}=\frac{\sqrt{1-x}}{\sqrt{2}}\\\\ <=> 2.2x=1-x\\\\ <=> 5x=1 <=> x=\frac{1}{5}[/tex]