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: \(2\cdot\sqrt{31}>2\cdot\sqrt{25}=2\cdot5=10\)

=>\(-2\sqrt{31}<-10\)

b: \(2\sqrt3-5-\left(\sqrt5-4\right)=2\sqrt3-\sqrt5-5+4=2\sqrt3-\sqrt5-1\)

\(\left(2\sqrt3\right)^2=12;\left(\sqrt5+1\right)^2=6+2\sqrt5\)

=>\(\left(2\sqrt3\right)^2-\left(\sqrt5+1\right)^2=12-6-2\sqrt5=6-2\sqrt5>0\)

=>\(\left(2\sqrt3-\sqrt5-1\right)\left(2\sqrt3+\sqrt5+1\right)>0\)

mà \(\left(2\sqrt3+\sqrt5+1\right)>0\)

nên \(\left(2\sqrt3-\sqrt5-1\right)>0\)

=>\(2\sqrt3-5-\left(\sqrt5-4\right)>0\)

=>\(2\sqrt3-5>\sqrt5-4\)

dell bt

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

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

\(\sqrt{2}\) + \(\sqrt{3}\)  > 2

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

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

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

\(3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}\)

\(2\sqrt{10}=\sqrt{2^2.10}=\sqrt{40}\)

thấy \(45>40=>\sqrt{45}>\sqrt{40}=>3\sqrt{5}>2\sqrt{10}\)

cảm ơn nha

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 \(4\sqrt{5}>5\sqrt{3}\)

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