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

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

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

\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)

Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)

Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)

\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)

Nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)

\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)

\(\sqrt{2}\) + \(\sqrt{3}\)  > 2

Sửa đề: \(S=\frac{4+\sqrt7}{3\sqrt2+\sqrt{4+\sqrt7}}+\frac{4-\sqrt7}{3\sqrt2-\sqrt{4-\sqrt7}}\)

\(=\frac{\sqrt2\left(4+\sqrt7\right)}{6+\sqrt{8+2\sqrt7}}+\frac{\sqrt2\left(4-\sqrt7\right)}{6-\sqrt{8-2\sqrt7}}\)

\(=\frac{\sqrt2\left(4+\sqrt7\right)}{6+\sqrt{\left(\sqrt7+1\right)^2}}+\frac{\sqrt2\left(4-\sqrt7\right)}{6-\sqrt{\left(\sqrt7-1\right)^2}}\)

\(=\frac{\sqrt2\left(4+\sqrt7\right)}{6+\sqrt7+1}+\frac{\sqrt2\left(4-\sqrt7\right)}{6-\left(\sqrt7-1\right)}=\frac{\sqrt2\left(4+\sqrt7\right)}{7+\sqrt7}+\frac{\sqrt2\left(4-\sqrt7\right)}{7-\sqrt7}\)

\(=\frac{1}{\sqrt2}\cdot\left\lbrack\frac{2\left(4+\sqrt7\right)}{\sqrt7\left(\sqrt7+1\right)}+\frac{2\left(4-\sqrt7\right)}{\sqrt7\left(\sqrt7-1\right)}\right\rbrack\)

\(=\frac{1}{\sqrt2}\cdot\left\lbrack\frac{8+2\sqrt7}{\sqrt7\left(\sqrt7+1\right)}+\frac{8-2\sqrt7}{\sqrt7\left(\sqrt7-1\right)}\right\rbrack\)

\(=\frac{1}{\sqrt2}\cdot\left\lbrack\frac{\left(\sqrt7+1\right)^2}{\sqrt7\left(\sqrt7+1\right)}+\frac{\left(\sqrt7-1\right)^2}{\sqrt7\left(\sqrt7-1\right)}\right\rbrack=\frac{1}{\sqrt2}\cdot\frac{\sqrt7+1+\sqrt7-1}{\sqrt7}=\frac{2\sqrt7}{\sqrt2\cdot\sqrt7}=\sqrt2\)

Ta có: \(R=\frac{3+\sqrt5}{2\sqrt2+\sqrt{3+\sqrt5}}+\frac{3-\sqrt5}{2\sqrt2-\sqrt{3-\sqrt5}}\)

\(=\frac{\sqrt2\left(3+\sqrt5\right)}{4+\sqrt{6+2\sqrt5}}+\frac{\sqrt2\left(3-\sqrt5\right)}{4-\sqrt{6-2\sqrt5}}\)

\(=\frac{\sqrt2\left(3+\sqrt5\right)}{4+\sqrt5+1}+\frac{\sqrt2\left(3-\sqrt5\right)}{4-\sqrt5+1}\)

\(=\frac{\sqrt2\left(3+\sqrt5\right)}{5+\sqrt5}+\frac{\sqrt2\left(3-\sqrt5\right)}{5-\sqrt5}=\sqrt2\cdot\left\lbrack\frac{3+\sqrt5}{\sqrt5\left(\sqrt5+1\right)}+\frac{3-\sqrt5}{\sqrt5\left(\sqrt5-1\right)}\right\rbrack\)

\(=\frac{\sqrt2}{2}\cdot\left\lbrack\frac{6+2\sqrt5}{\sqrt5\left(\sqrt5+1\right)}+\frac{6-2\sqrt5}{\sqrt5\left(\sqrt5-1\right)}\right\rbrack\)

\(=\frac{\sqrt2}{2}\cdot\left\lbrack\frac{\left(\sqrt5+1\right)^2}{\sqrt5\left(\sqrt5+1\right)}+\frac{\left(\sqrt5-1\right)^2}{\sqrt5\left(\sqrt5-1\right)}\right\rbrack=\frac{\sqrt2}{2}\cdot\frac{\sqrt5+1+\sqrt5-1}{\sqrt5}=\frac{\sqrt2}{2}\cdot2=\sqrt2\)

Do đó: R=S

b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)

mà 80>75

nên \(4\sqrt{5}>5\sqrt{3}\)

\(1.-3< -5+\sqrt{5}\)

\(2.-4>-2\sqrt{5}\)

\(3.-3\sqrt{5}< -6\)

2) \(4=\sqrt{16}\)

\(2\sqrt{5}=\sqrt{20}\)

mà 16<20

nên \(-4>-2\sqrt{5}\)

3) \(3\sqrt{5}=\sqrt{45}\)

\(6=\sqrt{36}\)

mà 45>36

nên \(-3\sqrt{5}< -6\)