a) \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\left|\sqrt{5}-2\right|-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\)

b) \(\left(4-\sqrt{7}\right)^2=4^2-2.4.\sqrt{7}+\sqrt{7}^2=16-8\sqrt{7}+7=23-8\sqrt{7}\)

c)  \(\sqrt{23+8\sqrt{7}}=\sqrt{\left(4+\sqrt{7}\right)^2}=\left|4+\sqrt{7}\right|=\sqrt{7}+4\)

a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)

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

=-4

b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)

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

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

c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)

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

\(=3\sqrt{5}-6\)

d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)

\(=\sqrt{7}-2+4-\sqrt{7}+8\)

=10

1)d) \(\sqrt{23+8\sqrt{7}}-\sqrt{7}\)

\(=\sqrt{4^2+2.4.\sqrt{7}+\sqrt{7^2}}-\sqrt{7}\)

\(=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)

\(=4+\sqrt{7}-\sqrt{7}\)

\(=4\)

a) \(\sqrt{9-4\sqrt{5}}+\sqrt{5}\)

=\(\sqrt{\left(\sqrt{2}\right)^2-2.2\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{5}\)

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

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

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

=\(\sqrt{2}\)

a) \(9+4\sqrt{5}=4+4\sqrt{5}+5=2^2+2\cdot2\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{5}+2\right)^2\left(ĐPCM\right)\)

a) \(9+4\sqrt{5}=\left(\sqrt{5}\right)^2+2.\sqrt{5}.2+2^2=\left(\sqrt{5}+2\right)^2\left(đpcm\right)\)

b)\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\left(đpcm\right)\)

c)\(\left(4-\sqrt{7}\right)^2=16-8\sqrt{7}+7=23-8\sqrt{7}\left(đpcm\right)\)

d)\(\sqrt{23+8\sqrt{7}}-\sqrt{7}=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}=4+\sqrt{7}-\sqrt{7}=4\left(đpcm\right)\)

- Đề sai nhiều vậy sửa lại đi bạn ;-;

e) Ta có: \(\sqrt{1-12x+36x^2}=5\)

\(\Leftrightarrow\left|6x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=6\\6x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{2}{3}\right\}\)

a) Ta có: \(9+4\sqrt{5}\)

\(=5+2\cdot\sqrt{5}\cdot2+4\)

\(=\left(\sqrt{5}+2\right)^2\)(đpcm)

b) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

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

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

=-2(ddpcm)

c) Ta có: \(\left(4-\sqrt{7}\right)^2\)

\(=16-2\cdot4\cdot\sqrt{7}+7\)

\(=23-8\sqrt{7}\)(đpcm)

d) Ta có: \(\sqrt{17-12\sqrt{2}}+2\sqrt{2}\)

\(=\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}+2\sqrt{2}\)

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

\(=3-2\sqrt{2}+2\sqrt{2}=3\)(đpcm)

\(a.VT=4+4\sqrt{5}+5=2^2+4\sqrt{5}+\sqrt{5}^2=\left(2+\sqrt{5}\right)^2=VP\)

\(b.\) Dựa vào câu a ta có: \(9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)

\(VT=\left|\sqrt{5}-2\right|-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2=VP\)

\(c.VT=16-8\sqrt{7}+7=4^2-8\sqrt{7}+\sqrt{7}^2=\left(4-\sqrt{7}\right)^2=VP\)

\(d.\) 

Ta có: \(17-12\sqrt{2}=8-12\sqrt{2}+9=\left(2\sqrt{2}\right)^2-12\sqrt{2}+3^2=\left(2\sqrt{2}-3\right)^2\)

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

\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)

\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)

\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)

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

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

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

\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

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

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

\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)