Ta có: \(X=\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)

<=> \(X^2=6-3\sqrt{2+\sqrt{3}}+2+\sqrt{2+\sqrt{3}}-2\sqrt{3}.\sqrt{4-\left(2+\sqrt{3}\right)}\)

<= \(X^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3}.\sqrt{2-\sqrt{3}}\)

<=> \(X^2=8-\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{6}\left(\sqrt{3}-1\right)\)

<=> \(X^2=8-4\sqrt{2}\)

<=> \(X^2-8=-4\sqrt{2}\)

=> \(X^4-16X+64=32\)

<=> \(X^4-16X^2+32=0\)

Vậy X là nghiệm phương trình \(X^4-16X^2+32=0\)

\(G=\sqrt{8\sqrt{3}-4\sqrt{6}-4\sqrt{2}+18}-\sqrt{14-4\sqrt{6}}\)

\(G=\sqrt{12+4+2-4\sqrt{6}+8\sqrt{3}-4\sqrt{2}}-\sqrt{12+4-4\sqrt{6}}\)

\(G=\left(2\sqrt{3}+2-\sqrt{2}\right)-\left(2\sqrt{3}-2\right)=4-\sqrt{2}\)

a)\(G=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\)

\(=\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)

\(=\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)

\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)

\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)

\(=\frac{2}{x+\sqrt{x}+1}\)

b) \(x+\sqrt{x}+1>0\Rightarrow G>0\)

\(x+\sqrt{x}+1>0+0+1=1\)

\(\Rightarrow\frac{2}{x+\sqrt{x}+1}< \frac{2}{1}=2\Rightarrow G< 2\)

\(\Rightarrow O< G< 2\)

ĐKXĐ: \(x\ge0\)

Áp dụng BĐT Cauchy cho 2 số dương: \(x+4\ge2\sqrt{4x}=4\sqrt{x}\)

\(\Rightarrow\frac{4\sqrt{x}}{x+4}\le1\Leftrightarrow T\le1\)

Dấu "=" xảy ra \(\Leftrightarrow x=4\)

ĐKXĐ: \(x\ge0,x\ne1\)

\(F=\frac{\sqrt{x}+1}{\sqrt{x}-1}< 0\Leftrightarrow\)\(\sqrt{x}+1\)và \(\sqrt{x}-1\)trái dấu 

\(\Rightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Vậy \(0< x< 1\)thỏa mãn đề bài.

a) đk: \(x\ge0\)

Ta có: \(M< -\frac{1}{2}\)

\(\Leftrightarrow-\frac{3}{\sqrt{x}+3}+\frac{1}{2}< 0\)

\(\Leftrightarrow\frac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\)

Mà \(2\left(\sqrt{x}+3\right)>0\left(\forall x\right)\Rightarrow\sqrt{x}-3< 0\)

\(\Leftrightarrow\sqrt{x}< 3\Rightarrow x< 9\)

Vậy \(0\le x< 9\)

b) Ta có: \(M=\frac{-3}{\sqrt{x}+3}\ge-\frac{3}{0+3}=-1\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\sqrt{x}=0\Rightarrow x=0\)

Vậy Min(M) = -1 khi x = 0