Đặt \(A=\sqrt{3+\sqrt{5}}\)

\(A\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)

\(A=\frac{\sqrt{5}+1}{2}\)\(< \frac{\sqrt{10}+1}{2}< \sqrt{10}+1\)

Vậy....

a) \(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)

b) \(3\sqrt{5}=\sqrt{45}>\sqrt{27}\)

c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{9}}< \sqrt{\dfrac{54}{9}}=6=\sqrt{\dfrac{150}{25}}=\dfrac{1}{5}\sqrt{150}\)

d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{4}}=\sqrt{\dfrac{3}{2}}< \sqrt{\dfrac{36}{2}}=6\sqrt{\dfrac{1}{2}}\)

\(A=\frac{\sqrt{\sqrt{3}+2}-\sqrt{-\sqrt{3}+2}}{\sqrt{\sqrt{3}+2}+\sqrt{-\sqrt{3}+2}}\)

\(A=\frac{\sqrt{2}}{\sqrt{2}}\cdot\frac{\sqrt{\sqrt{3}+2}-\sqrt{-\sqrt{3}+2}}{\sqrt{\sqrt{3}+2}+\sqrt{-\sqrt{3}+2}}\)

\(\Rightarrow A=\frac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{3}+1+\sqrt{3}-1}=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

\(\Rightarrow\frac{\sqrt{\sqrt{3}+2}-\sqrt{-\sqrt{3}+2}}{\sqrt{\sqrt{3}+2}+\sqrt{-\sqrt{3}+2}}=\frac{\sqrt{3}}{3}\)

ta tính VT ra rồi so sánh với VP

a,Ta có:

  \(\left(\sqrt{24}+\sqrt{45}\right)^2=24+45=69\)

\(12^2=144\)

Do 69<144 nên ...

b,tương tự ý a

a ) Ta co \(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}=5+7=12\)

vay \(\sqrt{24}+\sqrt{45}< 12\)

b)ta co \(\sqrt{37}-\sqrt{15}>\sqrt{4}-\sqrt{0}=2-0=2\)

vay \(\sqrt{37}-\sqrt{15}>2\)

a ) \(2\sqrt{5}-5\) và \(\sqrt{5}-3\)

Ta có ; \(2\sqrt{5}-5-\left(\sqrt{5}-3\right)\)

         \(=\sqrt{5}-8\)

         \(=\sqrt{5}-\sqrt{64}< 0\)

\(\Rightarrow2\sqrt{5}-5< \sqrt{5}-3\)

Vậy .................

b ) \(\sqrt{17}+\sqrt{26}\) và 9 

Ta có : 

\(\sqrt{17}>\sqrt{16}\)

\(\sqrt{26}>\sqrt{25}\)

\(\Rightarrow\sqrt{17}+\sqrt{26}>\sqrt{16}+\sqrt{25}=4+5=9\)

Vậy ...

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Mysterious Person

\(-2\sqrt{5}=-\sqrt{2^2.5}=-\sqrt{20}\)

\(-5\sqrt{2}=-\sqrt{5^2.2}=-\sqrt{50}\)

\(\Rightarrow-\sqrt{20}>-\sqrt{50}\)

hay \(-2\sqrt{5}>-5\sqrt{2}\)

> 

nha b hien

\(-2\sqrt{5}=-\sqrt{2^2.5}=-\sqrt{20}\)

\(-5\sqrt{2}=-\sqrt{5^2.2}=-\sqrt{50}\)

vì \(20< 50\)nen \(-\sqrt{20}>-\sqrt{50}\)