Ta có: (a-b)²\(\ge0\)

Suy ra (a+b)²\(\ge4ab\)

Suy ra \(\frac{1}{a+b}\)\(\le\frac{a+b}{4ab}\)

Suy ra \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(a,\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right).\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)

\(=2\)

\(b,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

Ta có \(0< 3y^2+1< 4y^2+4\)

=> \(y^4< y^4+3y^4+1< \left(y^2+2\right)^2\)

=> \(y^4< x^4< \left(y^2+2\right)^2\)

Mà x,y nguyên

=> \(x^2=y^2+1\)

=> \(y^4+2y^2+1=y^4+3y^2+1\)

=> \(y=0\)=> x=0

Vậy (x,y)=(0;0)

ĐK \(x\ge-2004\)

\(x^2-2004=-\sqrt{x+2004}\)

Đặt \(\sqrt{x+2004}=a\left(a\ge0\right)\)

=> \(\hept{\begin{cases}x^2-2004=-a\\a^2-2004=x\end{cases}}\)

=> \(\left(x+a\right)\left(x-a\right)+\left(a+x\right)=0\)

=> \(\left(x+a\right)\left(x-a+1\right)=0\)

=> \(\orbr{\begin{cases}x=-a\\x=a-1\end{cases}}\)

+ x=-a

=> \(x=-\sqrt{x+2004}\)

=> \(\hept{\begin{cases}x\le0\\x^2-x-2004=0\end{cases}}\)=> \(x=\frac{1-\sqrt{8017}}{2}\)(TmĐK)

+ \(x=a-1\)

=> \(x+1=-\sqrt{x+2004}\)

=> \(\hept{\begin{cases}x\le-1\\x^2+x-2003=0\end{cases}}\)=> \(x=\frac{-1-\sqrt{8013}}{2}\)(TTMĐK)

Vậy \(S=\left\{\frac{-1-\sqrt{8013}}{2};\frac{1-\sqrt{8017}}{2}\right\}\)