\(K=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}-\sqrt{\left(-8\right)^2}\)

    \(=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}-\sqrt{\left(-8\right)^2}\)

    \(=\sqrt{81-17}-8=\sqrt{64}-8=8-8=0\)

\(=\sqrt{81-17}-8\)

=8-8

=0

Đặt \(A=\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}\)

\(\Leftrightarrow A^2=18-2\sqrt{\left(9+\sqrt{17}\right)\left(9-\sqrt{17}\right)}\)

\(=18-2\sqrt{81-17}=2\)

\(\Rightarrow A=\sqrt{2}\)

\(\Rightarrow C=A-\sqrt{2}=0\)

a) \(H=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)

\(=\sqrt{81-17}=\sqrt{64}=8\)

b) \(K=\left(\sqrt{20}-3\sqrt{5}+\sqrt{80}\right).\sqrt{5}\)

\(=\sqrt{20}.\sqrt{5}-3\sqrt{5}.\sqrt{5}+\sqrt{80}.\sqrt{5}\)

\(=\sqrt{100}-3.5+\sqrt{400}=\sqrt{10^2}-15+\sqrt{20^2}\)

\(=10-15+20=15\)

\(H=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)   

\(=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)   

\(=\sqrt{9^2-\left(\sqrt{17}\right)^2}\)   

\(=\sqrt{81-17}\)   

\(=\sqrt{64}=8\)   

\(K=\left(\sqrt{20}-3\sqrt{5}+\sqrt{80}\right)\cdot\sqrt{5}\)   

\(=\sqrt{20}\cdot\sqrt{5}-3\sqrt{5}\cdot\sqrt{5}+\sqrt{80}\cdot\sqrt{5}\)   

\(=\sqrt{20\cdot5}-3\sqrt{5\cdot5}+\sqrt{80\cdot5}\)   

\(=\sqrt{100}-3\sqrt{25}+\sqrt{400}\)   

\(=10-3\cdot5+20\)   

\(=15\)

\(H=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}=\sqrt{81-17}=\sqrt{64}=8\)

\(K=\left(2\sqrt{5}-3\sqrt{5}+4\sqrt{5}\right)\sqrt{5}=3\sqrt{5}.\sqrt{5}=3.5=15\)

Đây là câu trả lời của mình, bạn chỉ cần áp dụng kĩ năng tính toán cơ bản là ra, học kĩ kiến thức cơ bản nhé

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

a: Ta có: \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{3}-\sqrt{5}-1\)

\(=\sqrt{3}-1\)

b: Ta có: \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)

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

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

c: Ta có: \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)

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

\(=4\sqrt{2}-1\)

a)

\(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{1}+1}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}-\sqrt{1}\\ =\sqrt{3}-\sqrt{1}\)

b)

\(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\\ =\sqrt{9-2\sqrt{9}\cdot\sqrt{8}+8}+\sqrt{18+2\sqrt{18}\cdot\sqrt{1}+1}\\ =\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =3-2\sqrt{2}+3\sqrt{2}+1\\ =4+\sqrt{2}\)

c)

\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{8-2\sqrt{8}\cdot\sqrt{4}+4}+\sqrt{8+2\sqrt{8}\cdot\sqrt{1}+1}\\ =\sqrt{\left(2\sqrt{2}-2\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =2\sqrt{2}-2+2\sqrt{2}+1\\ =4\sqrt{2}-1\)

Nếu đề đúng:

Sử dụng liên hợp để trục căn thức ở mẫu:

\(\frac{1}{\sqrt{1}+\sqrt{5}}=\frac{\sqrt{5}-1}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{\sqrt{5}-1}{5-1}=\frac{\sqrt{5}-1}{4}\) 

Tương tự như vậy ta sẽ có:

\(N=\frac{\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}+\frac{\sqrt{13}-\sqrt{9}}{\left(\sqrt{13}-\sqrt{9}\right)\left(\sqrt{13}+\sqrt{9}\right)}+\frac{\sqrt{17}-\sqrt{13}}{\left(\sqrt{17}-\sqrt{13}\right)\left(\sqrt{17}+\sqrt{13}\right)}\)

\(+\frac{\sqrt{21}-\sqrt{17}}{\left(\sqrt{21}-\sqrt{17}\right)\left(\sqrt{21}+\sqrt{17}\right)}+\frac{\sqrt{25}-\sqrt{23}}{\left(\sqrt{25}-\sqrt{23}\right)\left(\sqrt{25}+\sqrt{23}\right)}\)

\(=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+\frac{\sqrt{17}-\sqrt{13}}{4}+\frac{\sqrt{21}-\sqrt{17}}{4}+\frac{\sqrt{25}-\sqrt{23}}{4}\)

\(=\frac{\sqrt{5}-1+\sqrt{13}-\sqrt{9}+\sqrt{17}-\sqrt{13}+\sqrt{21}-\sqrt{17}+\sqrt{25}-\sqrt{23}}{4}\)

\(=\frac{\sqrt{5}-1-\sqrt{9}+\sqrt{21}+\sqrt{25}-\sqrt{23}}{4}=\frac{\sqrt{5}-1-3+\sqrt{21}+5-\sqrt{23}}{4}=\frac{1+\sqrt{5}+\sqrt{21}-\sqrt{23}}{4}\)

a, \(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)

\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{5-8}}}\)

\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)

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

\(=\sqrt{10+2\left(\sqrt{5}-2\right)}\)

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

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

b, \(=\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)

\(=\sqrt{6+2\sqrt{5-\left(2\sqrt{3}+1\right)}}\)

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

\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)

\(=\sqrt{6+2\sqrt{3}-2}\)

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

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

\(=\sqrt{3^2+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}-\sqrt{3^2-2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)

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

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

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

\(=4\sqrt{2}\)

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

\(=\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}-\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)

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

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

\(=3+2\sqrt{2}-3+2\sqrt{2}=4\sqrt[]{2}\)