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.

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

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

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

Xem lại câu c) và d) 

b: =căn 10-3+4-căn 10=1

a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)

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

a) \(\sqrt{11+4\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)

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

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

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

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

a) \(\sqrt{11+4\sqrt{7}}-\sqrt{11-4\sqrt{7}}=\sqrt{\left(2+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-2\right)^2}=2+\sqrt{7}-\sqrt{7}+2=4\)

b) \(A=\sqrt{11-4\sqrt{6}}-\sqrt{11+4\sqrt{6}}\)

\(\Rightarrow A^2=11-4\sqrt{6}-2\sqrt{\left(11-4\sqrt{6}\right)\left(11+4\sqrt{6}\right)}+11+4\sqrt{6}\)

\(A^2=22-2\sqrt{121-96}\)

\(A^2=22-2\sqrt{25}=22-2.5=12\)

\(\Rightarrow A=-\sqrt{12}\)(Chú ý \(A< 0\))

b) \(\sqrt{11-4\sqrt{6}}-\sqrt{11+4\sqrt{6}}\)

\(=\sqrt{11-2.2\sqrt{2}.\sqrt{3}}-\sqrt{11+2.2\sqrt{2}.\sqrt{3}}\)

= \(\sqrt{8-2.2\sqrt{2}.\sqrt{3}+3}-\sqrt{8+2.2\sqrt{2}.\sqrt{3}+3}\)

\(=\sqrt{\left(\sqrt{8}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{8}+\sqrt{3}\right)^2}\)

\(=\sqrt{8}-\sqrt{3}-\sqrt{8}-\sqrt{3}=-2\sqrt{3}\)

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

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

=3-4=-1

b: \(=\sqrt{6+4\sqrt{2}}-\sqrt{11-2\sqrt{18}}\)

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

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

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

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

\(=4\sqrt{5}\)

e) \(\sqrt{x^2}=\left|-8\right|\Rightarrow\left|x\right|=8\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

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

\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}+\sqrt{2}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}+\sqrt{2}\)

\(=\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}+\sqrt{2}=\dfrac{\sqrt{7}-1}{\sqrt{2}}-\dfrac{\sqrt{7}+1}{\sqrt{2}}+\sqrt{2}\)

\(=-\dfrac{2}{\sqrt{2}}+\sqrt{2}=-\sqrt{2}+\sqrt{2}=0\)

f) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)

\(=\sqrt{\dfrac{12+2\sqrt{11}}{2}}-\sqrt{\dfrac{12-2\sqrt{11}}{2}}+3\sqrt{2}\)

\(=\sqrt{\dfrac{\left(\sqrt{11}\right)^2+2.\sqrt{11}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{11}\right)^2-2.\sqrt{11}.1+1^2}{2}}+3\sqrt{2}\)

\(=\sqrt{\dfrac{\left(\sqrt{11}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{11}-1\right)^2}{2}}+3\sqrt{2}\)

\(=\dfrac{\left|\sqrt{11}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{11}-1\right|}{\sqrt{2}}+3\sqrt{2}=\dfrac{\sqrt{11}+1}{\sqrt{2}}-\dfrac{\sqrt{11}-1}{\sqrt{2}}+3\sqrt{2}\)

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

\(\sqrt{\left|4\sqrt{6}-11\right|}-\sqrt{4\sqrt{6}+11}\)

Vì \(4\sqrt{6}< 11\) nên khi thoát dấu GTTĐ, ta được:

\(\sqrt{11-4\sqrt{6}}-\sqrt{11+4\sqrt{6}}\)

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

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

=|√3-2√2|-|√3+2√2|

= 2√2-√3-√3-2√2

= -2√3

\(\sqrt{\left|4\sqrt{6}-11\right|}-\sqrt{4\sqrt{6}+11}\)

Ta có:

\(4\sqrt{6}< 11\)

\(\Rightarrow\sqrt{11-4\sqrt{6}}-\sqrt{11+4\sqrt{6}}\)

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

Từ đây rút gọn căn của 2 bên rồi tính nốt

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

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

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

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

6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)

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