Ta có: 

\(R=\)\(\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

\(=\)\(\dfrac{\sqrt{10}+3\sqrt{2}}{5+\sqrt{5}}+\dfrac{\sqrt{10}-3\sqrt{2}}{5-\sqrt{5}}\)

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

\(=\dfrac{4\sqrt{2}}{4\sqrt{5}}=\sqrt{\dfrac{2}{5}}\)

Làm câu S tương tự như này rồi đối chiếu kết quả nha

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

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

Vậy A = B 

A = 11 

B = 7 

--> A > B 

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

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

=> \(A\sqrt{2}=B\sqrt{2}\) => A = B

a) \(\sqrt[3]{7+5\sqrt{2}}=\sqrt{2}+1\)

b) \(-6\sqrt[3]{7}=\sqrt[3]{\left(-6\right)^3\cdot7}=\sqrt[3]{-1512}\)

\(7\sqrt[3]{-6}=\sqrt[3]{7^3\cdot\left(-6\right)}=\sqrt[3]{-2058}\)

mà -1512>-2058

nên \(-6\sqrt[3]{7}>7\cdot\sqrt[3]{-6}\)

\(A=\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{5}+1-\sqrt{5}=1\)

\(B=\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)

Do đó: A=B

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

\(\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}\right)^3+1^3+3.2+3\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)

--> Bằng nhau

Mình chọn nhầm lớp 8 chứ thật ra câu hỏi ở bên lớp 9 

a) Ta có \(5=\sqrt{25}\)

Vì \(\sqrt{25}>\sqrt{11}\) nên \(5>\sqrt{11}\)

b) Ta có \(4=\sqrt{16}\)

Vì \(\sqrt{13}< \sqrt{16}\) nên \(\sqrt{13}< 4\)

c) Ta có \(-7=-\sqrt{49}\)

Vì \(-\sqrt{49}< -\sqrt{43}\) nên \(-7< -\sqrt{43}\)

d) Ta có \(-5=-\sqrt{25}\)

Vì \(-\sqrt{21}>-\sqrt{25}\) nên \(-\sqrt{21}>-5\)