\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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

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

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

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

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

\(A=\left(\dfrac{\sqrt{12}-\sqrt{6}}{2\sqrt{2}-2}-\dfrac{6\sqrt{6}}{3}\right).\dfrac{1}{\sqrt{6}}\)

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

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

\(A=-\sqrt{6}.\dfrac{1}{\sqrt{6}}\)

\(A=-1\)

a: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)

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

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

b: \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

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

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

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

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

\(=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}\)

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

d: \(\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}\)

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

\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\left(4-\sqrt{3}\right)+3}\)

\(=\sqrt{16-3+3}=\sqrt{16}=4\)

e: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)

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

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

`A=\sqrt{6-2\sqrt{5}}`

`A=\sqrt{(\sqrt{5}-1)^2}`

`A=\sqrt{5}-1`

_________

`B=\sqrt{4-\sqrt{12}}=\sqrt{4-2\sqrt{3}}`

`B=\sqrt{(\sqrt{3}-1)^2}`

`B=\sqrt{3}-1`

_________

`C=\sqrt{19-8\sqrt{3}}`

`C=\sqrt{(4-\sqrt{3})^2}`

`C=4-\sqrt{3}`

_________

`D=\sqrt{5-2\sqrt{6}}`

`D=\sqrt{(\sqrt{3}-\sqrt{2})^2}`

`D=\sqrt{3}-\sqrt{2}`

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

\(B=\sqrt{4-\sqrt{12}}=\sqrt{4-\sqrt{4.3}}=\sqrt{4-2\sqrt{3}}=\sqrt{\sqrt{3^2}-2\sqrt{3}+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

\(C=\sqrt{19-8\sqrt{3}}=\sqrt{19-2.4.\sqrt{3}}\sqrt{\sqrt{3}^2-2.4.\sqrt{3}+4^2}=\sqrt{\left(\sqrt{3}-4\right)^2}=\sqrt{3}-4\)

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

Lời giải:
a.

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

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

$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$

b.

$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$

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

$=|\sqrt{3}-3|+|\sqrt{3}+3|$

$=(3-\sqrt{3})+(\sqrt{3}+3)=6$

c.

$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$

$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$

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

1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

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

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

\(A=\sqrt{5}-1-\sqrt{5}-1\)

\(A=-2\)

\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

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

\(B=4\)

Sửa đề :

\(C=\sqrt{14-6\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)

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

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

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

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

b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}=3+\sqrt{2}-3+\sqrt{2}\)

\(=2\sqrt{2}\)

c) \(x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(x-4\right)^2}=x-4+\left|x-4\right|\)

\(=x-4+x-4\left(x>4\right)=2x-8\)

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

e) \(\dfrac{x^2+2\sqrt{2}x+2}{x+\sqrt{2}}=\dfrac{\left(x+\sqrt{2}\right)^2}{x+\sqrt{2}}=x+\sqrt{2}\)

g) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{1}{\sqrt{2}}\)

a) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)

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

=-1

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

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

\(=2\sqrt{2}\)

c) Ta có: \(x-4+\sqrt{x^2-8x+16}\)

\(=x-4+x-4=2x-8\)

d) Ta có: \(\dfrac{x^2-5}{x+\sqrt{5}}\)

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

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

1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

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

\(=\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)

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

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

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

\(=4\)

2) \(\sqrt{12-6\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)

\(=\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{3^2+2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

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

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

\(=6\)