\(A=2\sqrt{6}\)

\(B=2\sqrt{4}=4\)

\(C=2\sqrt{7}\)

Bn lm rõ ra chút đc k ak

a,Ta có :  \(1-\sqrt{3}\); \(\sqrt{2}-\sqrt{6}=\sqrt{2}\left(1-\sqrt{3}\right)\Rightarrow1-\sqrt{3}< \sqrt{2}\left(1-\sqrt{3}\right)\)

Vậy \(1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

b, Đặt A =  \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)(*)

\(\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)

\(=\sqrt{7}+1-\sqrt{7}+1-2=0\Rightarrow A=0\)

Vậy (*) = 0 

1: 

Ta có: \(\sqrt{2}-\sqrt{6}\)

\(=\sqrt{2}\left(1-\sqrt{3}\right)< 0\)

\(\Leftrightarrow1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)

b: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)

a: \(\frac{3}{4+\sqrt{9+4\sqrt5}}\)

\(=\frac{3}{4+\sqrt{\left(\sqrt5+2\right)^2}}\)

\(=\frac{3}{4+\sqrt5+2}=\frac{3}{6+\sqrt5}=\frac{3\left(6-\sqrt5\right)}{36-5}=\frac{3\left(6-\sqrt5\right)}{31}\)

b: \(\frac{\sqrt3}{\sqrt2+\sqrt{5+2\sqrt6}}\)

\(=\frac{\sqrt3}{\sqrt2+\sqrt{\left(\sqrt3+\sqrt2\right)^2}}=\frac{\sqrt3}{\sqrt2+\sqrt3+\sqrt2}\)

\(=\frac{\sqrt3}{2\sqrt2+\sqrt3}=\frac{\sqrt3\left(2\sqrt2-\sqrt3\right)}{8-3}=\frac{2\sqrt6-3}{5}\)

c: \(\frac{3}{\sqrt5+\sqrt7-\sqrt2}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{\left(\sqrt5+\sqrt7\right)^2-2}\)

\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{10+2\sqrt{35}}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(\sqrt{35}+5\right)\left(\sqrt{35}-5\right)}\)

\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(35-25\right)}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{20}\)

\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{225}}\)

\(\dfrac{1}{\sqrt{k}}=\dfrac{2}{\sqrt{k}+\sqrt{k}}< \dfrac{2}{\sqrt{k+1}+\sqrt{k}}\\ =\dfrac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)

\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{225}}\\ < 2\left(\sqrt{226}-\sqrt{225}\right)+2\left(\sqrt{225}-\sqrt{224}\right)+...+2\left(\sqrt{3}-\sqrt{2}\right)\\ =2\left(\sqrt{226}-\sqrt{225}+\sqrt{225}-\sqrt{224}+...+\sqrt{3}-\sqrt{2}\right)\\ =2\left(\sqrt{226}-\sqrt{2}\right)< 2\left(\sqrt{225}-\sqrt{2}\right)< 2\left(\sqrt{225}-\sqrt{1}\right)=28\left(đpcm\right)\)

Vậy \(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{225}}< 28\)

​\(a,\sqrt{9}-4\sqrt{5}-\sqrt{5}=\sqrt{3^2}-4\sqrt{5}-\sqrt{5}=3-5\sqrt{5}\)​

\(b,\sqrt{3}-2\sqrt{2}-\sqrt{3}+2\sqrt{2}=0\)

\(c,\sqrt{11}-6\sqrt{2}+3+\sqrt{2}=\sqrt{11}-5\sqrt{2}+3\)

\(a,\sqrt{9}-4\sqrt{5}-\sqrt{5}=3-3\sqrt{5}\)

\(b,\sqrt{3}-2\sqrt{2}-\sqrt{3}+2\sqrt{2}=0\)