a.\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\)

b.\(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2\)

c.\(\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3\)

d.\(\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)

1)\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{26^2}=\sqrt{5}-2+26=24-\sqrt{5}\)

2) \(=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)

3) \(=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)\(=\left[{}\begin{matrix}1\left(x>1\right)\\-1\left(x< 1\right)\end{matrix}\right.\)

4) \(=\sqrt{\left(\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}=\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}=2\sqrt{\dfrac{1}{2}}=\sqrt{2}\)

2. \(\dfrac{x^2-5}{x+\sqrt{5}}=\dfrac{x^2-\left(\sqrt{5}\right)^2}{x+\sqrt{5}}=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)

3. \(\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{\sqrt{x^2-2.x.1+1^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{|x-1|}{x-1}=\left[{}\begin{matrix}x-1>0\left(x>1\right)\\x-1< 0\left(x< 1\right)\end{matrix}\right.=\left[{}\begin{matrix}=1\\=\dfrac{x+1}{x-1}\end{matrix}\right.\)

\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

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

\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)

\(=2+\sqrt{3}-2+\sqrt{3}\)

\(=2\sqrt{3}\)

\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

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

\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)

\(=3+\sqrt{2}-3+\sqrt{2}\)

\(=2\sqrt{2}\)

\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

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

\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)

\(=3+2\sqrt{2}+3-2\sqrt{2}\)

\(=6\)

\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)

\(=2+\sqrt{5}-\sqrt{5}+2\)

\(=4\)

\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)

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

\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)

\(=1+\sqrt{5}-\sqrt{5}+1\)

\(=2\)

\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

\(A=\sqrt{3}+2+2-\sqrt{3}\)

A = 2 + 2

A = 4

\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

\(B=\sqrt{2}+3+3-\sqrt{2}\)

B = 3 + 3

B = 6

\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

\(C=3+2\sqrt{2}+3-2\sqrt{2}\)

C = 3 + 3

C = 6

\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(D=\sqrt{5}+2-\sqrt{5}+2\)

D = 2 + 2

D = 4

\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)

\(E=\sqrt{5}+1-\sqrt{5}+1\)

E = 1 + 1

E = 2

a: \(=2\sqrt{2}+1-3=2\sqrt{2}-2\)

b: \(=\sqrt{3}+1-2\sqrt{3}-1=-\sqrt{3}\)

c: \(=2-\sqrt{3}+\sqrt{3}-1=1\)

a: \(=\sqrt{8+2\cdot2\sqrt{2}\cdot\sqrt{5}+5}+\sqrt{8-2\cdot2\sqrt{2}\cdot\sqrt{5}+5}\)

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

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

b: \(=2\cdot\sqrt{17-3\sqrt{32}}\)

\(=2\cdot\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}\)

\(=2\left(3-2\sqrt{2}\right)=6-4\sqrt{2}\)

A= \(\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\)\(1+\sqrt{7}+\sqrt{7}-1=2\sqrt{7}\)

\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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

\(A=-\sqrt{2}-\sqrt{1}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+....-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}\)

\(A=\sqrt{9}-\sqrt{1}=3-1=2\)

với n >0, ta có :

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

Gọi biểu thức đã cho là A

\(A=\frac{1}{-\left(\sqrt{2}-\sqrt{1}\right)}-\frac{1}{-\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{1}{-\left(\sqrt{8}-\sqrt{7}\right)}-\frac{1}{-\left(\sqrt{9}-\sqrt{8}\right)}\)

\(A=-\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-...-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{9}-\sqrt{8}}\)

\(A=-\left(\sqrt{2}+\sqrt{1}\right)+\left(\sqrt{3}+\sqrt{2}\right)-...-\left(\sqrt{8}+\sqrt{7}\right)+\left(\sqrt{9}+\sqrt{8}\right)\)

\(A=-\sqrt{1}+\sqrt{9}=2\)

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