\(P=\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(\Leftrightarrow Px+P\sqrt{x}+P-\sqrt{x}-2=0\)

\(\Leftrightarrow Px+\sqrt{x}\left(P-1\right)+P-2=0\)
\(\Leftrightarrow Pt^2+\left(P-1\right)t+\left(P-2\right)=0\)(Với \(t=\sqrt{x},t\ge0\))
\(\Delta=\left(P-1\right)^2-4P\left(P-2\right)=-3P^2+6P+1\ge0\Leftrightarrow\frac{3-2\sqrt{3}}{3}\le P\le\frac{3+2\sqrt{3}}{3}\)
Do P nguyên nên P={0;1;2}
Nếu P=0 =>Vô nghiệm
Nếu P=1=> Không tm đề bài
Nếu P=2 => x=0 (Không tm x>0)
Vậy................................
 

\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)

\(=\sqrt{1}=1\)

b,c

\(\sqrt{13+4\sqrt{3}}=\sqrt{13+2\sqrt{12}}=\sqrt{12}+1=2\sqrt{3}+1\)

=>BT=\(\sqrt{5-\left(2\sqrt{3}+1\right)}+\sqrt{3+\left(2\sqrt{3}+1\right)}\)

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

\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)

c,\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)

\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

\(B=x-4\sqrt{x}\)

\(B=x-2.\sqrt{x}.2+4-4\)

\(B=\left(\sqrt{x}-2\right)^2-4\)

\(Vì\left(\sqrt{x}-2\right)^2\ge0\Rightarrow B=\left(\sqrt{x}-2\right)^2-4\ge-4\)

\(\text{Dấu "=" xảy ra }\Leftrightarrow\sqrt{x}-2=0\)

\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=\sqrt{2}\)

\(\text{Vậy Min B=-4}\Leftrightarrow x=\sqrt{2}\)

\(VT=\sqrt{6-2\sqrt{5}}-\sqrt{5}\)

\(=\sqrt{\left(\sqrt{5}-1\right) ^2}-\sqrt{5}\)

\(=\left|\sqrt{5}-1\right|-\sqrt{5}\)

\(=\sqrt{5}-1-\sqrt{5}\)

\(=-1=VP\)  (đpcm)