\(A=\dfrac{2}{\sqrt{17}+\sqrt{15}}\) ; \(B=\dfrac{2}{\sqrt{15}+\sqrt{13}}\)

Mà \(\sqrt{17}+\sqrt{15}>\sqrt{15}+\sqrt{13}>0\)

\(\Rightarrow\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{15}+\sqrt{13}}\)

\(\Rightarrow A< B\)

\(A=\sqrt{17}-\sqrt{15}=\dfrac{2}{\sqrt{17}+\sqrt{15}}\)

\(B=\sqrt{15}-\sqrt{13}=\dfrac{2}{\sqrt{13}+\sqrt{15}}\)

mà \(\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{13}+\sqrt{15}}\)

nên A<B

So sánh theo cách của lớp 6 tích:

\(\sqrt{12}=3,464101615\)

\(\sqrt{17}=4,123105626\)

\(\Rightarrow\sqrt{12}< \sqrt{17}\)

Ta có :

\(\sqrt{12}< \sqrt{17}\)

Vì \(\sqrt{12}< \sqrt{16}< \sqrt{17}\)

\(\Rightarrow\sqrt{12}< \sqrt{17}\)

ta có \(17>12\Rightarrow\sqrt{17}>\sqrt{12}\)

Giúp mình với mn

a: \(1< \sqrt{2}\)

nên \(2< \sqrt{2}+1\)

b: \(2\sqrt{31}=\sqrt{124}\)

\(10=\sqrt{100}\)

mà 124>100

nên \(2\sqrt{31}>10\)

c: \(-3\sqrt{11}=-\sqrt{99}\)

\(-\sqrt{12}=-\sqrt{12}\)

mà 99>12

nên \(-3\sqrt{11}< -\sqrt{12}\)

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 \(4\sqrt{5}>5\sqrt{3}\)

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

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

\(\Rightarrow A< B\)

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

a ) 

\(\sqrt{31}+4< \sqrt{36}+4=10\left(1\right)\)

\(6+\sqrt{17}>6+\sqrt{16}=6+4=10\left(2\right)\)

Từ ( 1 ) ; ( 2 ) 

\(\Rightarrow\sqrt{31}+4< 10< 6+\sqrt{17}\)

\(\Rightarrow\sqrt{31}+4< \sqrt{17}+6\)

b ) 

\(\sqrt{3}+\sqrt{2}>\sqrt{1}+\sqrt{1}=2\)

c ) 

\(\sqrt{12+13}=\sqrt{25}=5\left(1\right)\)

\(\sqrt{12}+\sqrt{13}>\sqrt{4}+\sqrt{9}=2+3=5\left(2\right)\)

Từ ( 1 ) ; ( 2 ) 

\(\Rightarrow\sqrt{12+13}< \sqrt{12}+\sqrt{13}\)

:V

khó vc

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

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

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