= \(\sqrt{12-2.2\sqrt{3}.5+25}-\sqrt{12+2.2\sqrt{3}.5+25}\)

= \(\sqrt{\left(2\sqrt{3}-5\right)^2}-\sqrt{\left(2\sqrt{3}+5\right)^2}\)

= \(|2\sqrt{3}-5|-2\sqrt{3}-5\)

=\(5-2\sqrt{3}-2\sqrt{3}-5=-4\sqrt{3}\)

bây giờ vẫn còn công chúa

\(=\sqrt{\left(5-2\sqrt{3}\right)^2}+\sqrt{\left(5+2\sqrt{3}\right)^2}\)

\(=\left|5-2\sqrt{3}\right|+\left|5+2\sqrt{3}\right|\)

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

\(=10\)

\(2\sqrt[]{37+20\sqrt[]{3}}-\sqrt[]{73-40\sqrt[]{3}}\)

\(=2\sqrt[]{25+2.5.2\sqrt[]{3}+12}-\sqrt[]{48-2.5.4\sqrt[]{3}+25}\)

\(=2\sqrt[]{\left(5+2\sqrt[]{3}\right)^2}-\sqrt[]{\left(5-4\sqrt[]{3}\right)^2}\)

\(=2\left|5+2\sqrt[]{3}\right|-\left|5-4\sqrt[]{3}\right|\)

\(=2\left(5+2\sqrt[]{3}\right)-\left(4\sqrt[]{3}-5\right)\left(vì.4\sqrt[]{3}>5\right)\)

\(=10+4\sqrt[]{3}-4\sqrt[]{3}+5\)

\(=15\)

\(A=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}-\sqrt{37-20\sqrt{3}}\)

\(=\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}-\sqrt{\left(5-2\sqrt{3}\right)^2}\)

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

\(=2\sqrt{3}-1\)

\(\sqrt{13-4\sqrt{3}}-\sqrt{37-20\sqrt{3}}\)

\(=\sqrt{12-4\sqrt{3}+1}-\sqrt{25-20\sqrt{3}+12}\)

\(=\sqrt{\left(2\sqrt{3}-1\right)^2}-\sqrt{\left(5-2\sqrt{3}\right)^2}\)

\(=\left|2\sqrt{3}-1\right|-\left|5-2\sqrt{3}\right|\)

\(=2\sqrt{3}-1-5+2\sqrt{3}\)

\(=4\sqrt{3}-6\)

\(\left(5+2\sqrt{3}\right)\cdot\sqrt{37-20\sqrt{3}}\\ =\left(5+2\sqrt{3}\right)\cdot\sqrt{25-2\cdot10\sqrt{3}+12}\\ =\left(5+2\sqrt{3}\right)\cdot\sqrt{5^2-2\cdot5\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2}\\ =\left(5+2\sqrt{3}\right)\cdot\sqrt{\left(5-2\sqrt{3}\right)^2}\\ =\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)\\ =5^2-\left(2\sqrt{3}\right)^2\\ =25-12=13\)

a) \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\)

b) \(\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}=\dfrac{\sqrt{3}+1-\left(\sqrt{3}-1\right)}{3-1}=1\)

c) \(2\sqrt{5}-3\sqrt{45}+\sqrt{500}=2\sqrt{5}-9\sqrt{5}+10\sqrt{5}=3\sqrt{5}\)

d) \(\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\sqrt{3}=\dfrac{1-\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\dfrac{1-3-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-2-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=-\sqrt{2}\)

e) \(\dfrac{1}{2+\sqrt{3}}-\dfrac{1}{2-\sqrt{3}}+5\sqrt{3}=\dfrac{2-\sqrt{3}-\left(2+\sqrt{3}\right)}{4-3}+5\sqrt{3}=-2\sqrt{3}+5\sqrt{3}=3\sqrt{3}\)

f) \(\sqrt{3}-\sqrt{4+2\sqrt{3}}=\sqrt{3}-\left(\sqrt{3}+1\right)=-1\)

g) \(\dfrac{5-\sqrt{5}}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\sqrt{5}-\dfrac{4}{\sqrt{5}+1}=\dfrac{5+\sqrt{5}-4}{\sqrt{5}+1}=1\)

h)\(\sqrt{37-20\sqrt{3}+\sqrt{37+20\sqrt{3}}}=\sqrt{37-20\sqrt{3}+\left(5+2\sqrt{3}\right)}=\sqrt{42-18\sqrt{3}}=\sqrt{\left(3\sqrt{3}+3\right)^2+6}\)

Bài 1 : 

Ta có : 

\(\sqrt{37-20\sqrt{3}}+\sqrt{37+20\sqrt{3}}=\sqrt{25-2.5.2\sqrt{3}+12}\)

\(+\sqrt{25+2.5.2\sqrt{3}+12}\)

\(=\sqrt{\left(5-2\sqrt{3}\right)^2}+\sqrt{\left(5+2\sqrt{3}\right)^2}\)

\(5-2\sqrt{3}+5+2\sqrt{3}\)

\(=5+5=10\)

Bài 2 : 

Với x , y , z > 0 . Ta có : 

+ ) \(\frac{x}{y}+\frac{y}{x}\ge2\left(1\right)\)

+ ) \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\left(2\right)\)

+ ) \(x^2+y^2+z^2\ge xy+yz+zx\Leftrightarrow\frac{x^2+y^2+z^2}{xy+yz+zx}\ge1\left(3\right)\)

Xảy ra đăng thức ở : \(\left(1\right),\left(2\right),\left(3\right)\Leftrightarrow x=y=z\) . Ta có : 

\(P=\frac{ab+bc+ca}{a^2+b^2+c^2}+\left(a+b+c\right)^2.\frac{\left(a+b+c\right)}{abc}\)

\(=\frac{ab+bc+ca}{a^2+b^2+c^2}+\left(a^2+b^2+c^2+2ab+2bc+2ca\right).\frac{\left(a+b+c\right)}{abc}\)

Áp dụng các bất đẳng thức (1) , (2) , (3) ta được : 

\(P\ge\frac{ab+bc+ca}{a^2+b^2+c^2}+\left(a^2+b^2+c^2\right).\frac{9}{ab+bc+ca}+2.9\)

\(=\left(\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{a^2+b^2+c^2}{ab+bc+ca}\right)+8.\frac{a^2+b^2+c^2}{ab+bc+ca}+18\)

\(\ge2+8+18=28\)

Dấu " = "  xảy ra \(\Leftrightarrow\hept{\begin{cases}a^2+b^2+c^2=ab+bc+ca\\ab=bc=ca\end{cases}\Leftrightarrow a=b=c}\)