a. ĐKXĐ: x\geq 0 ; x khác 1
P=[tex](\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}).(\frac{1-x}{\sqrt{2}})^2[/tex]
P=[tex](\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}).(\frac{1-x}{\sqrt{2}})^2[/tex]
P=[tex](\frac{(\sqrt{x}-2)(\sqrt{x}+1)-(\sqrt{x}+2)(\sqrt{x}-1)}{(x-1)(\sqrt{x}+1)}).(\frac{1-x}{\sqrt{2}})^2[/tex]
P=[tex](\frac{-2\sqrt{x}}{(x-1)(\sqrt{x}+1)}).(\frac{1-x}{\sqrt{2}})^2[/tex]
P=[tex]\frac{2.\sqrt{x}(1-x)}{\sqrt{x}+1}[/tex]
P=[tex] 2. \sqrt{x}.(1-\sqrt{x})[/tex]
b. Với 0<x<1 thì 1-x >0
[tex]\sqrt{x}[/tex] >0
Nên P > 0
c. Áp dụng bđt Côsi ta có
[tex]4.\sqrt{x}(1-\sqrt{x}) \leq (\sqrt{x}+1-\sqrt{x})^2[/tex]
\Rightarrow P \leq [tex]\frac{1}{2}[/tex]
Dấu = xảy ra \Leftrightarrow [tex]x=\frac{1}{4}[/tex]