\(C=\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right).\left(8-2\sqrt{10+2\sqrt{5}}\right)}=\sqrt{\left(8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2\right)=\sqrt{64-4\left(10+2\sqrt{5}\right)}}\)

\(C=\sqrt{64-40-8\sqrt{5}}=\sqrt{24-8\sqrt{5}}\)

\(C=\sqrt{20-2.2.2\sqrt{5}+4}=\sqrt{\left(2\sqrt{5}-2\right)^2}\)

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

2:

ĐKXĐ: x>=3

 \(\Leftrightarrow\sqrt{x-3+2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}+\sqrt{x-3-2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}=2\sqrt{3}\)

=>\(\left|\sqrt{x-3}+\sqrt{3}\right|+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)

\(\Leftrightarrow\sqrt{x-3}+\sqrt{3}+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)

\(\Leftrightarrow\sqrt{x-3}+\left|\sqrt{x-3}-\sqrt{3}\right|=\sqrt{3}\)(1)

TH1: x>=6

(1) trở thành \(\sqrt{x-3}+\sqrt{x-3}-\sqrt{3}=\sqrt{3}\)

=>\(2\sqrt{x-3}=2\sqrt{3}\)

=>x-3=3

=>x=6(nhận)

TH2: 3<=x<6

Phương trình (1) sẽ là;

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

=>\(\sqrt{3}=\sqrt{3}\)(luôn đúng)

1:

\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)

\(=16+2\cdot\sqrt{64-4\cdot\left(10+2\sqrt{5}\right)}\)

\(=16+2\cdot\sqrt{24-8\sqrt{5}}\)

\(=16+2\cdot\sqrt{20-2\cdot2\sqrt{5}\cdot2+4}\)

\(=16+2\cdot\sqrt{\left(2\sqrt{5}-2\right)^2}\)

\(=16+2\cdot\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)

\(=10+2\cdot\sqrt{10}\cdot\sqrt{2}+2\)

\(=\left(\sqrt{10}+\sqrt{2}\right)^2\)

=>\(A=\sqrt{10}+\sqrt{2}\)

binh phuong len di ban

\(\frac{5}{\sqrt{2}-7}-\frac{4}{3\sqrt{2}+5}-\frac{7}{4-5\sqrt{2}}\)

\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2.\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right)\left(8-2\sqrt{10+2\sqrt{5}}\right)}\)

\(A^2=16+2.\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}=16+2.\sqrt{24-8\sqrt{5}}=16+4.\sqrt{6-2\sqrt{5}}\)

\(A^2=16+4.\sqrt{\left(\sqrt{5}-1\right)^2}=16+4.\left(\sqrt{5}-1\right)=12+4\sqrt{5}\)

=> A = \(\sqrt{12+4\sqrt{5}}=\sqrt{2}\sqrt{6+2\sqrt{5}}=\sqrt{2}.\left(\sqrt{5}+1\right)=\sqrt{10}+\sqrt{2}\)

\(\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}\\ =\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{2}.\sqrt{4}-\sqrt{2}.\sqrt{5}}{2-\sqrt{5}}\\ =\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}\\ =\dfrac{1}{\sqrt{2}+1}-\sqrt{2}\\ =\dfrac{1-\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\\ =\dfrac{1-2-\sqrt{2}}{\sqrt{2}+1}\\ =\dfrac{-\sqrt{2}-1}{\sqrt{2}+1}\\ =\dfrac{-\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\\ =-1\)

\(\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}=-1+\sqrt{2}-\sqrt{2}=-1\)