Bạn xem hộ mk đề cậu b nhé căn 5- căn 2 hay là căn 5 - 2

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

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: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)

1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)

\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)

\(=6-8=-2\)

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

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

\(=9-5=4\)

3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)

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

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

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

4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)

=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn

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

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

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

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

\(=22-2\sqrt{5}\)

=> \(A=\sqrt{22-2\sqrt{5}}\)

6) Đặt \(B=\sqrt{10+2\sqrt{11}}-\sqrt{10-2\sqrt{11}}\)

\(\Leftrightarrow B^2=\left(\sqrt{10+2\sqrt{11}}-\sqrt{10-2\sqrt{11}}\right)^2\)

\(=10+2\sqrt{11}-2\sqrt{\left(10+2\sqrt{11}\right)\left(10-2\sqrt{11}\right)}+10-2\sqrt{11}\)

\(=20-2\sqrt{100-44}\)

\(=20-4\sqrt{14}\)

=> \(B=\sqrt{20-4\sqrt{14}}\)

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

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

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

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

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

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