b, \(M=A-B=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\left(\frac{5}{x+\sqrt{x}-6}+\frac{1}{\sqrt{x}-2}\right)\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}-\frac{5}{x+\sqrt{x}-6}-\frac{1\left(\sqrt{x}+3\right)}{x+\sqrt{x}-6}\)

\(=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+3\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)

\(ab+bc+ca=1\)\(\Rightarrow\)\(\hept{\begin{cases}a+b+c\ge\sqrt{3}\\a^2+b^2+c^2\ge1\end{cases}}\)

\(\left(a-\frac{1}{\sqrt{3}}\right)^2\ge0\)\(\Leftrightarrow\)\(a\le\frac{\sqrt{3}}{2}a^2+\frac{\sqrt{3}}{6}\)

\(P=\Sigma\frac{a^2\left(1-2b\right)^2}{b\left(1-2b\right)}\ge\frac{\left(a+b+c-2\right)^2}{\left(a+b+c\right)-2\left(a^2+b^2+c^2\right)}\ge\frac{\left(a+b+c-2\right)^2}{\frac{\sqrt{3}-4}{2}\Sigma a^2+\frac{\sqrt{3}}{2}}\ge\sqrt{3}-2\)

\(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\)

\(VP=\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\le\frac{a+b}{2}\sqrt{2\left(a+b\right)}\)\(\Rightarrow\)\(VP^2\le\frac{\left(a+b\right)^3}{2}\) (1) 

chứng minh bổ đề: \(VT^2=\left(\frac{\left(a+b\right)^2}{2}+\frac{a+b}{4}\right)^2\ge\frac{\left(a+b\right)^3}{2}\)

\(\Leftrightarrow\)\(\frac{\left(a+b\right)^4}{4}+\frac{\left(a+b\right)^2}{16}+\frac{\left(a+b\right)^3}{4}\ge\frac{\left(a+b\right)^3}{2}\)

\(\Leftrightarrow\)\(\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge\left(a+b\right)^3\)

Có: \(\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge2\sqrt{\frac{\left(a+b\right)^6}{4}}=\left(a+b\right)^3\)\(\Rightarrow\)\(VT^2\ge\frac{\left(a+b\right)^3}{2}\) (2) 

(1) và (2) => \(VT^2\ge VP^2\) => \(VT\ge VP\) ( đpcm ) 

ta có P=\(\frac{x^2}{x\sqrt{y+3}}+\frac{y^2}{y\sqrt{z+3}}+\frac{z^2}{z\sqrt{x+3}}\ge\frac{\left(x+y+z\right)^2}{x\sqrt{y+3}+y\sqrt{z+3}+z\sqrt{x+3}}\)

mà \(\left(x\sqrt{y+3}+...\right)^2\le\left(x+y+z\right)\left(xy+yz+zx+3x+3y+3z\right)\le3\left(9+3\right)=36\) ( vì xy+yz+zx<=3)

=>\(x\sqrt{y+3}+...\le6\Rightarrow P\ge\frac{9}{6}=\frac{3}{2}\)

dấu = xảy ra <=> x=y=z=1

Có: \(2a=AE+AF+EF=AE+AF+\sqrt{AE^2+AF^2}\ge\sqrt{AE.AF}\left(2+\sqrt{2}\right)\)

\(\Leftrightarrow\)\(\sqrt{AE.AF}\le\frac{2a}{2+\sqrt{2}}\)\(\Leftrightarrow\)\(\frac{1}{2}AE.AF=\frac{a^2}{3+2\sqrt{2}}\) không đổi 

Dấu "=" xảy ra khi \(AE=AF\)\(\Rightarrow\)\(\hept{\begin{cases}2AE+EF=2a\\2AE^2=EF^2\end{cases}}\Leftrightarrow\hept{\begin{cases}EF=2a-2AE\\EF=\sqrt{2}AE\end{cases}}\)

\(\Rightarrow\)\(2a-2AE=\sqrt{2}AE\)\(\Leftrightarrow\)\(AE=\frac{2a}{2+\sqrt{2}}\)

D= \(\frac{x^2-2x+2000}{x^2}\)= 1-\(\frac{2x}{x^2}\)+\(\frac{2000}{x^2}\)

Đặt t= \(\frac{1}{2}\)=> D= 2000t²-2t+1 = (\(20\sqrt{5}t\))²-2.\(20\sqrt{5}t\).\(\frac{1}{20\sqrt{5}}\)+\(\left(\frac{1}{20\sqrt{5}}\right)^2\)\(-\left(\frac{1}{20\sqrt{5}}\right)^2\)+1

D= (\(20\sqrt{5}t\)-\(\frac{1}{20\sqrt{5}}\)) ²+\(\frac{1999}{2000}\)\(\ge\)\(\frac{1999}{2000}\)

Min D= \(\frac{1999}{2000}\)khi \(20\sqrt{5}t\)\(-\frac{1}{20\sqrt{5}}\)= 0 => t = \(\frac{1}{2000}\)=> \(\frac{1}{x}\)= \(\frac{1}{2000}\)=> x= 2000