\(\sqrt{\left(3+2\sqrt{5}\right)^2}-\sqrt{\left(3-2\sqrt{5}\right)^2}\\ =3+2\sqrt{5}-2\sqrt{5}+3\left(vi2\sqrt{5}>3\right)\\ =6\)

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

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

 Do \(3+2\sqrt{5}>0\Rightarrow\left|3+2\sqrt{5}\right|=3+2\sqrt{5}\)

\(3-2\sqrt{5}< 0\Rightarrow\left|3-2\sqrt{5}\right|=2\sqrt{5}-3\)

\(=3+2\sqrt{5}-\left(2\sqrt{5}-3\right)=3+2\sqrt{5}+3-2\sqrt{5}=6\)

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

\(=\frac{\sqrt{4}}{3+\sqrt{5}}+\frac{\sqrt{4}}{3-\sqrt{5}}=\frac{2.\left(3-\sqrt{5}\right)+2.\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\frac{12}{4}=3\)

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

= \(\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{9-5}}\)+ \(\sqrt{\frac{\left(3+\sqrt{5}\right)^2}{9-5}}\)

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

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

= \(\frac{6}{2}\)

=3

#mã mã#

a: \(A=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

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

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

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

b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

\(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)

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

\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)

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

\(=-\dfrac{2}{\sqrt{x}+1}\)

c: Khi x=9 thì \(B=\dfrac{-2}{\sqrt{9}+1}=\dfrac{-2}{3+1}=-\dfrac{2}{4}=-\dfrac{1}{2}\)

d: |B|=A

=>\(\left|-\dfrac{2}{\sqrt{x}+1}\right|=4\)

=>\(\dfrac{2}{\sqrt{x}+1}=4\) hoặc \(\dfrac{2}{\sqrt{x}+1}=-4\)

=>\(\sqrt{x}+1=\dfrac{1}{2}\) hoặc \(\sqrt{x}+1=-\dfrac{1}{2}\)

=>\(\sqrt{x}=-\dfrac{1}{2}\)(loại) hoặc \(\sqrt{x}=-\dfrac{3}{2}\)(loại)

Lời giải:
Gọi biểu thức là A

\(A=\left[3-\frac{\sqrt{5}(\sqrt{5}-1)}{1-\sqrt{5}}\right]\left[\frac{\sqrt{5}(\sqrt{2}+\sqrt{3})}{\sqrt{2}+\sqrt{3}}-3\right]\)

\(=[3-\frac{-\sqrt{5}(1-\sqrt{5})}{1-\sqrt{5}}](\sqrt{5}-3)=(3--\sqrt{5})(\sqrt{5}-3)=(3+\sqrt{5})(\sqrt{5}-3)=5-3^2=-4\)

Ta có: \(\left(3-\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}+\sqrt{3}}-3\right)\)

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

=5-9

=-4

1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)

\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)

\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)

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

\(=-8\sqrt{3}\)

2) \(A=\sqrt{12-4x}\) có nghĩa khi:

\(12-4x\ge0\)

\(\Leftrightarrow4x\le12\)

\(\Leftrightarrow x\le\dfrac{12}{4}\)

\(\Leftrightarrow x\le3\)

3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)

\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)

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

\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)

Ta có:

\(x=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}}\)   ( x> 0 )

\(\Rightarrow x^2=6+2\sqrt{\left(3+\sqrt{5+2\sqrt{3}}\right)\left(3-\sqrt{5+2\sqrt{3}}\right)}\)

\(=6+2\sqrt{9-5-2\sqrt{3}}\)

\(=6+2\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=6+2\sqrt{3}-2=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)

\(\Rightarrow x=\sqrt{3}+1\)

Vậy :

\(A=x^2-2x-2=4+2\sqrt{3}-2\sqrt{3}-2-2\)

\(=0\)

b: Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\left(x+\sqrt{x}+1+\sqrt{x}\right)\)

\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}-1}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)