\(A=\sqrt{12+\sqrt{12+\sqrt{12}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}< \sqrt{12+\sqrt{12+\sqrt{16}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{9}}}}\)\(=7\)

\(B=\sqrt{14}+\sqrt{11}>\sqrt{13,69}+\sqrt{10,89}=7\)

\(\Rightarrow A< B\)

Ta có:

 \(12< 16\Rightarrow\sqrt{12}< \sqrt{16}=4\\ 6< 9\Rightarrow\sqrt{6}< \sqrt{9}=3\)

\(\Rightarrow A< \sqrt{12+\sqrt{12+4}}+\sqrt{6+\sqrt{6+\sqrt{6+3}}}=\sqrt{12+4}+\sqrt{6+3}=4+3=7\) (1)

Lại có :

\(B=\sqrt{14}+\sqrt{11}\Rightarrow B^2=25+2\sqrt{14.11}=25+2\sqrt{154}>25+2\sqrt{144}=25+2.12=49=7^2\)

Mà B > 0

\(\Rightarrow B>7\) (2)

Từ (1),(2) suy ra A<B

    3.991546341         >                   2.997403267

chuc 1 bạn học tốt

\(6< 9\Rightarrow\sqrt{6}< \sqrt{9}=3\)

\(\Rightarrow\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}< \sqrt{6+\sqrt{6+\sqrt{6+3}}}\)

\(=\sqrt{6+\sqrt{6+3}}\)\(=\sqrt{6+3}\)\(=3\)

\(12>9\Rightarrow\sqrt{12}>\sqrt{9}=3\)

\(\Rightarrow\sqrt{12+\sqrt{12+\sqrt{12}}}>\sqrt{12}>3\)

\(\Rightarrow\sqrt{12+\sqrt{12+\sqrt{12}}}>\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}\)

Lời giải:
\(2\sqrt{12}>2\sqrt{9}=2.3=6>3\)

\(\sqrt{6}> \sqrt{5}\)

\(\Rightarrow 2\sqrt{12}+\sqrt{6}> 3+\sqrt{5}\)

ghi de sai ban oi

\(A=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\)

\(A< \sqrt{2,25}+\sqrt{6,25}+\sqrt{12,25}+\sqrt{20,25}+\sqrt{30,25}+\sqrt{42,25}=24=B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

Ta có: \(12>9\)

\(6\sqrt{3}>4\sqrt{5}\)

Do đó: \(12+6\sqrt{3}>9+4\sqrt{5}\)

\(\Leftrightarrow\sqrt{12+6\sqrt{3}}>\sqrt{9+4\sqrt{5}}\)

A=√2+√6+√12+√20+√30+√42

A= 23.7579

B= 24

vậy => B > A

giúp mk với

A=√2+√6+√12+√20+√30+√42

A=23,75790715

Mà B=24

=>A<B