\(S=x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)\)\(=\left(x^2+y^2\right)\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]-x^2y^2\left(x+y\right)\)

\(=\left[\left(\frac{\sqrt{2}+1}{2}\right)^2+\left(\frac{\sqrt{2}-1}{2}\right)^2\right]\)\(\cdot\left[\left(\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}\right)^3-3\frac{\sqrt{2}+1}{2}\frac{\sqrt{2}-1}{2}\left(\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2-1}}{2}\right)\right]\)\(-\left(\frac{\sqrt{2}+1}{2}\right)^2\left(\frac{\sqrt{2}-1}{2}\right)^2\left(\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}\right)\)

\(=\frac{\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}-1\right)^2}{4}\cdot\left[\left(\sqrt{2}\right)^3-\frac{3}{4}\sqrt{2}\right]\)\(-\frac{\left(\sqrt{2}+1\right)^2\left(\sqrt{2}-1\right)^2}{4\cdot4}\sqrt{2}\)

\(=\frac{2+2\sqrt{2}+1+2-2\sqrt{2}+1}{4}\left(2\sqrt{2}-\frac{3\sqrt{2}}{4}\right)-\frac{1}{16}\sqrt{2}\)

\(=\frac{6}{4}\cdot\frac{5\sqrt{2}}{4}-\frac{\sqrt{2}}{16}=\frac{30\sqrt{2}-\sqrt{2}}{16}=\frac{29\sqrt{2}}{16}\)

b) \(B=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2+2}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

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

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

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

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

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

Đây là câu trả lời của mình. Bài này là bài dễ, bạn nên xem lại kiến thức cơ bản ở bài trục căn ở mẫu nhé!

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

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

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

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

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

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

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

\(=\sqrt{81-17}\)   

\(=\sqrt{64}=8\)   

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

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

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

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

\(=10-3\cdot5+20\)   

\(=15\)