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-\(\sqrt{11}\)

\(=\frac{\left(\sqrt{13}+\sqrt{11}\right)^2+\left(\sqrt{13}-\sqrt{11}\right)^2}{\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{13}+\sqrt{11}\right)}\)

\(=\frac{13+2\sqrt{143}+11+13-2\sqrt{143}+11}{13-11}\)

\(=\frac{48}{2}=24\)

a: \(=2\cdot\sqrt{\dfrac{18-2\sqrt{77}}{4}}-\sqrt{20+6\sqrt{11}}\)

\(=\sqrt{11}-\sqrt{7}-\sqrt{11}-3=-\sqrt{7}-3\)

b: B=\(=\left(\sqrt{13}-1\right)\cdot\sqrt{\dfrac{7+\sqrt{13}}{18}}+\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

Đặt \(C=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(\Leftrightarrow C^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{16-10-2\sqrt{5}}\)

\(=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)

=>\(C=\sqrt{5}+1\)

\(B=\left(\sqrt{13}-1\right)\cdot\sqrt{\dfrac{14+2\sqrt{13}}{36}}+\sqrt{5}+1\)

\(=\dfrac{\left(\sqrt{13}-1\right)\left(\sqrt{13}+1\right)}{6}+\sqrt{5}+1\)

=(13-1)/6+căn5+1

=3+căn5

a) \(x^2+8=3\sqrt{x^3+8}\)

\(\left(x^2+8\right)^2=\left(3\sqrt{x^2+8}\right)^2\)

\(x^4+16x^2+64=9x^2+72\)

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