* \(\sqrt{2}\)A = \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+\sqrt{14}=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{14}=\sqrt{7}-1-\left(\sqrt{7}+1\right)+\sqrt{14}=\sqrt{14}-2\)

=> A = \(\sqrt{7}-\sqrt{2}\)

* B là 6,5 hay 6*5 vậy bạn

nếu 6,5 thì : B cũng nhân \(\sqrt{2}\) biểu thức trở thành

\(\sqrt{2}B=\sqrt{13+2\sqrt{12}}+\sqrt{13-2\sqrt{12}}+4\sqrt{3}=\sqrt{\left(1+\sqrt{12}\right)^2}+\sqrt{\left(\sqrt{12}-1\right)^2}+4\sqrt{3}=1+\sqrt{12}+\sqrt{12}-1+4\sqrt{3}=4\sqrt{3}+4\sqrt{3}=8\sqrt{3}\)

=> B = \(\dfrac{8\sqrt{3}}{\sqrt{2}}=4\sqrt{6}\)

nếu 6*5 thì : bạn tách hai căn đầu thành một biểu thức rồi bình phương lên rồi giải , sau đó trục căn , biểu thức luôn dương nhé , mấy bài này nếu không thể tách thì làm cách này cũng được

* C thì mik chỉ bít pt được nhiu đây thôi , bạn thông cảm nhé\(\sqrt{29-6\sqrt{20}}=\sqrt{\left(\sqrt{20}-3\right)^2}=\sqrt{20}+3=2\sqrt{5}-3\)

* D = \(\sqrt{13-2\cdot2\sqrt{2}\cdot\sqrt{5}}-\sqrt{53+2\cdot2\sqrt{2}\cdot3\sqrt{5}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{2}+3\sqrt{5}\right)^2}=2\sqrt{2}-\sqrt{5}-2\sqrt{2}-3\sqrt{5}=-4\sqrt{5}\)

Câu C có sai đề ko? Tui sửa đây!

Ta có: \(C=\sqrt{46+6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

=> \(C=\sqrt{45+2.3\sqrt{5}+1}-\sqrt{20-2.3.2\sqrt{5}+9}\)

=> \(C=\sqrt{\left(3\sqrt{5}+1\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)

=> \(C=\left|3\sqrt{5}+1\right|-\left|2\sqrt{5}-3\right|\)

=> \(C=3\sqrt{5}+1-2\sqrt{5}+3=4+\sqrt{5}\)

1.

$\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3}}-\sqrt{3+1-2\sqrt{3}}$

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

$=|\sqrt{3}+1|-|\sqrt{3}-1|=2$

2.

\(\sqrt{12+6\sqrt{3}+\sqrt{12-6\sqrt{3}}}=\sqrt{12+6\sqrt{3}+\sqrt{9+3-2\sqrt{9.3}}}=\sqrt{12+6\sqrt{3}+\sqrt{(3-\sqrt{3})^2}}\)

\(=\sqrt{12+6\sqrt{3}+3-\sqrt{3}}=\sqrt{15+5\sqrt{3}}\)

3.

\(\sqrt{9-4\sqrt{2}+\sqrt{9+4\sqrt{2}}}=\sqrt{9-4\sqrt{2}+\sqrt{8+1+2\sqrt{8.1}}}\)

\(=\sqrt{9-4\sqrt{2}+\sqrt{2\sqrt{2}+1)^2}}=\sqrt{9-4\sqrt{2}+2\sqrt{2}+1}=\sqrt{10-2\sqrt{2}}\)

4.

\(\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{9-\sqrt{32}}}}=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{8+1-2\sqrt{8.1}}}}\)

\(=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{(\sqrt{8}-1)^2}}}\) \(=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{8}-1}}=\sqrt{\sqrt{2}+2+\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{\sqrt{2}+2+\sqrt{(2+1+2\sqrt{2}}}=\sqrt{\sqrt{2}+2+\sqrt{(\sqrt{2}+1)^2}}=\sqrt{\sqrt{2}+2+\sqrt{2}+1}\)

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

5.

\(\sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9+2\sqrt{20.9}}}\)

\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}+3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}+3)}=\sqrt{3}\)

6.

\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\)

\(=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(=\sqrt{(2+5+2\sqrt{2.5})+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{2+5+2\sqrt{2.5}}\)

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

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

a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}}\)

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

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

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

\(=\sqrt{\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\)

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

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

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

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

b) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)

\(=\sqrt{6+2\sqrt{5}-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}\)

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

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

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

c) Sửa đề: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)

Ta có: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)

\(=\sqrt{2+\sqrt{5+\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\)

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

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

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

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

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

d) Ta có: \(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)

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

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

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

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

\(=\dfrac{8\sqrt{5}}{2\sqrt{2}}=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)

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.

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

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

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

f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)

mik sửa lại câu f , tí nhé :

f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

a,\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)

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

A=\(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)

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

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

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