\(\text{a, Ta có:}\)

\(3\sqrt{7}=\sqrt{3^27}=\sqrt{63}\)

\(9=\sqrt{81}\)

\(\text{Vì}:\sqrt{81}>\sqrt{63}\Rightarrow3\sqrt{7}< 9\)

\(\text{b, Vì}\) \(-\sqrt{3}>-\sqrt{5}\Rightarrow-\sqrt{\sqrt{3}}>-\sqrt{\sqrt{5}}\)

\(c,\sqrt{51}-\sqrt{3}\approx5,4>5\)

\(d.\text{Vì}\) \(5>\sqrt{5}\Rightarrow\sqrt{85+5}>\sqrt{85+\sqrt{5}}\)

1: \(8^2=64=22+32=22+2\cdot16=22+2\cdot\sqrt{256}\)

\(\left(\sqrt{8}+\sqrt{14}\right)^2=22+2\cdot\sqrt{112}\)

mà \(16>\sqrt{112}\)

nên 8^2>(căn 8+căn 14)^2

=>8>căn 8+căn 14

2: \(\left(2+\sqrt{3}\right)^2=7+4\sqrt{3}\)

\(\left(3+\sqrt{2}\right)^2=11+6\sqrt{2}\)

mà 7<11 và 4căn 3<6căn 2(48<72)

nên (2+căn 3)^2<(3+căn 2)^2

=>2+căn 3<3+căn 2

\(5\sqrt{2}+\sqrt{75}=5\sqrt{2}+5\sqrt{3}\)

\(5\sqrt{3}+\sqrt{50}=5\sqrt{3}+5\sqrt{2}\)

\(\Rightarrow5\sqrt{2}+\sqrt{75}=5\sqrt{3}+\sqrt{50}\)

1/ bình phương hai vế được (căn11)^2+(căn5)^2=11+5   4^2=16 vậy căn 11+căn 5=4

2/ tương tự (3 căn3 )^2=27   (căn19)^2-(căn 2)^2=19-2=17  vậy 3 căn 3 >căn 19-căn2

Lời giải:

$\sqrt{3}+5> \sqrt{1}+5=6$

$\sqrt{2}+\sqrt{11}< \sqrt{4}+\sqrt{16}=6$

$\Rightarrow \sqrt{3}+5> \sqrt{2}+\sqrt{11}$

b: \(\sqrt{\dfrac{3}{2}}>\sqrt{\dfrac{2}{2}}=1\)

a: \(\left(2\sqrt{5}-3\sqrt{2}\right)^2=38-12\sqrt{10}=1+37-12\sqrt{10}\)

\(1^2=1\)

mà \(37-12\sqrt{10}< 0\)

nên \(2\sqrt{5}-3\sqrt{2}< 1\)

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