Với n > 0 Ta có:

\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)

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

\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)

\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)

\(\sqrt{16}+\sqrt{9}=3+4=7\)

Gọi A= \(\sqrt{5-\sqrt{13+2\sqrt{11}}}\) - \(\sqrt{5+\sqrt{13+2\sqrt{11}}}\) 

Lấy A bình phương rồi áp dụng hằng đẳng thức số 2 sẽ ra:

A^2 = \(10-\) \(2\sqrt{25-\left(13+2\sqrt{11}\right)}\)

= \(10-2\sqrt{11-2\sqrt{11}+1}\)

= \(10-2\sqrt{\left(\sqrt{11}-1\right)^2}\)

= \(12-2\sqrt{11}\)

=\(11-2\sqrt{11}+1\)

= \(\left(\sqrt{11}-1\right)^2\)

Suy ra A= \(\sqrt{11}-1\)

\(a=\sqrt{5-\sqrt{13+2\sqrt{11}}}\); \(b=\sqrt{5+\sqrt{13+2\sqrt{11}}}\)dễ thấy \(a< b\)

ta có \(a^2+b^2=10;a.b=\left(\sqrt{11}-1\right)^{ }\).

Từ đây ta có \(\left(a-b\right)^2=\left(\sqrt{11}-1\right)^2\)kết hợp với a<b => a-b=1-\(\sqrt{11}\)

a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)

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

Thay x = 4 => \(\sqrt{x}=2\) vào B ta được : 

\(B=\frac{2+5}{2-3}=-7\)

b, Ta có : Với \(x\ge0;x\ne9\)

\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)

\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)

Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)