1. ĐKXĐ: \(x>0\)
\(A=\sqrt{x}+\frac{1}{\sqrt{x}}-1\ge2\sqrt{\frac{\sqrt{x}}{\sqrt{x}}}-1=2-1=1\)
\(A_{min}=1\) khi \(x=1\)
2. ĐKXĐ: \(x\ge0\)
\(x=\frac{4-2\sqrt{3}}{4}=\left(\frac{\sqrt{3}-1}{2}\right)^2\Rightarrow\sqrt{x}=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow P=\frac{2\left(\sqrt{3}-1\right)}{\left(\frac{\sqrt{3}-1}{2}+1\right)^2}=\frac{8\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)^2}=\frac{8\left(\sqrt{3}-1\right)^3}{4}=-20+12\sqrt{3}\)
\(P=\frac{1}{2}\Rightarrow\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)^2}=\frac{1}{2}\Leftrightarrow8\sqrt{x}=x+2\sqrt{x}+1\)
\(\Leftrightarrow x-6\sqrt{x}+1=0\Rightarrow\sqrt{x}=3\pm2\sqrt{2}\)
\(\Rightarrow x=17\pm12\sqrt{2}\)