1) \(A=\left(\sqrt{7-\sqrt{21}+4\sqrt{5}}\right)^2=7-\sqrt{21}+4\sqrt{5}\)

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

\(\Rightarrow A-B=1-\sqrt{21}+6\sqrt{5}=\left(1+\sqrt{180}\right)-\sqrt{21}>0\)

\(\Rightarrow A>B\Rightarrow\sqrt{7-\sqrt{21}+4\sqrt{5}}>\sqrt{5}-1\)

2) \(C=\left(\sqrt{5}+\sqrt{10}+1\right)^2=5+10+1+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}\)

\(=26+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}>26+10>35=\left(\sqrt{35}\right)^2\)

Vậy \(\sqrt{5}+\sqrt{10}+1>\sqrt{35}\)

3) \(\left(\frac{15-2\sqrt{10}}{3}\right)^2=\frac{225-60\sqrt{10}+40}{9}=\frac{265-60\sqrt{10}}{9}=\frac{265}{9}-\frac{20\sqrt{10}}{3}< 15\)

Vậy nên \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)

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a. \(\sqrt{35}+\sqrt{99}< \sqrt{36}+\sqrt{100}=6+10=16\)

\(\Rightarrow\sqrt{35}+\sqrt{99}< 16\)

b. \(\sqrt{24}< \sqrt{25}=5\)

\(\sqrt{5}+\sqrt{10}>\sqrt{4}+\sqrt{9}=2+3=5\)

\(\Rightarrow\sqrt{24}< \sqrt{5}+\sqrt{10}\)

\(\sqrt{3\sqrt{2}}=\sqrt{\sqrt{3^2\cdot2}}=\sqrt{\sqrt{18}}\)

\(\sqrt{2\sqrt{3}}=\sqrt{\sqrt{2^2\cdot3}}=\sqrt{\sqrt{12}}\)

từ trên ta suy ra

\(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)

Ta có:

\(\sqrt[3]{7}< \sqrt[3]{8}=2\) và \(\sqrt{15}< \sqrt{16}=4\), suy ra \(\sqrt[3]{7}+\sqrt{15}< 6\).

\(\sqrt{10}>\sqrt{9}=3\) và \(\sqrt[3]{28}>\sqrt[3]{27}=3\), suy ra \(\sqrt{10}+\sqrt[3]{28}>6\).

Vậy \(\sqrt[3]{7}+\sqrt{15}< \sqrt{10}+\sqrt[3]{28}\).