\(P=\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}+1\right):\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}-1\right)\) \(ĐKXĐ:x\ne1\)

\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+1\right):\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-1\right)\)

\(P=\left(\sqrt{x}+1\right):\left(\sqrt{x}-1\right)\)

\(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

b)  theo bài ra \(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

theo bài ra \(P< 1\)\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}< 1\)

\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}-1< 0\)

\(\Leftrightarrow\frac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\frac{2}{\sqrt{x}-1}< 0\)

\(\Rightarrow\sqrt{x}-1< 0\)

\(\Rightarrow\sqrt{x}< 1\)

\(\Rightarrow x< 1\) ( tm ĐK \(x\ne1\))

vậy \(x< 1\)thì \(P< 1\)

\(\sqrt{8-4\sqrt{3}}-\sqrt{8+4\sqrt{3}}\)

\(=\sqrt{8-2\sqrt{12}}-\sqrt{8+2\sqrt{12}}\)

\(=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{6}\right)^2}\)

\(=|\sqrt{6}-\sqrt{2}|-|\sqrt{2}+\sqrt{6}|\)

\(=\sqrt{6}-\sqrt{2}-\sqrt{2}-\sqrt{6}\)

\(=-2\sqrt{2}\)

Đặt \(\frac{1}{a}=x\); \(\frac{2}{b}=y;\frac{3}{c}=z\)

=>VT = \(\frac{z^3}{x^2+z^2}+\frac{x^3}{y^2+x^2}+\frac{y^3}{y^2+z^2}\)

Ta có \(\frac{z^3}{x^2+z^2}=z-\frac{x^2z}{x^2+z^2}\ge z-\frac{x^2z}{2xz}=z-\frac{x}{2}\)

CMTT: 

=> VT \(\ge\frac{x+y+z}{2}=\frac{3}{2}\). Dấu = khi a=1; b=2; z=3