$A=\sqrt[]{6-2\sqrt[]{5}}+\sqrt[]{6+2\sqrt[]{5}}$ 
$A=\sqrt[]{6-2\sqrt[]{5}.1}+\sqrt[]{6+2\sqrt[]{5}.1}$ 
$A=\sqrt[]{(\sqrt[]{5}-1)^2}+\sqrt[]{(\sqrt[]{5}+1)^2}$ 
$A=|\sqrt[]{5}-1|+|\sqrt[]{5}+1|$
$A=\sqrt[]{5}-1+\sqrt[]{5}+1$ 
$A=2\sqrt[]{5}$ 

$B=\sqrt[]{3+2\sqrt[]{2}}-\sqrt[]{6-4\sqrt[]{2}}$ 
$B=\sqrt[]{3+2\sqrt[]{2}.1}-\sqrt[]{6-2.2.\sqrt[]{2}}$ 
$B=\sqrt[]{(\sqrt[]{2}+1)^2}-\sqrt[](2-{\sqrt[]{2})^2}$ 
$B=|\sqrt[]{5}-1|+|\sqrt[]{5}+1|$
$B=\sqrt[]{2}+1-2+\sqrt[]{2}$ 
$B=2\sqrt[]{2}-1$ 

$C=\sqrt[]{6+2\sqrt[]{5-\sqrt[]{13+4\sqrt[]{3}}}}$
$C=\sqrt[]{6+2\sqrt[]{5-\sqrt[]{13+2.2\sqrt[]{3}.1}}}$ 
$C=\sqrt[]{6+2\sqrt[]{5-\sqrt[]{(2\sqrt[]{3}+1)^2}}}$ 
$C=\sqrt[]{6+2\sqrt[]{5-2\sqrt[]{3}-1}}$
$C=\sqrt[]{6+2\sqrt[]{4-2\sqrt[]{3}}}$
$C=\sqrt[]{6+2\sqrt[]{4-2.\sqrt[]{3}.1}}$
$C=\sqrt[]{6+2\sqrt[]{(\sqrt[]{3}-1)^2}}$
$C=\sqrt[]{6+2(\sqrt[]{3}-1)}$
$C=\sqrt[]{6+2\sqrt[]{3}-2}$
$C=\sqrt[]{4+2\sqrt[]{3}}$
$C=\sqrt[]{4+2.\sqrt[]{3}.1}$
$C=\sqrt[]{(\sqrt[]{3}+1)^2}$
$C=\sqrt[]{3}+1$ 

$D = \sqrt{4 + \sqrt{8}}.\sqrt{2 + \sqrt{2 + \sqrt{2}}}.\sqrt{2 - \sqrt{2 + \sqrt{2}}}$

 $D= \sqrt{4 + \sqrt{4.2}}.\sqrt{2^2 - \sqrt{2 + \sqrt{2}}^2}$

$D= \sqrt{4 + 2\sqrt{2}}.\sqrt{4 - (2 + \sqrt{2})}$

$D= \sqrt{2(2 + \sqrt{2})}.\sqrt{4 - 2 - \sqrt{2}}$

$D= \sqrt{2}.\sqrt{2 + \sqrt{2}}.\sqrt{2 - \sqrt{2}}$

$D= \sqrt{2}.\sqrt{2^2 - \sqrt{2}^2}$

$D= \sqrt{2}.\sqrt{4 - 2}$

$D= \sqrt{2}.\sqrt{2}$

$D= 2$

@Holliwood#: @ TEAM_IQ_2000#:

Đáp án:

a) A= 2√5

b) B= -1 +2√2

c) C= 1+√3

d) D= 2

Giải thích các bước giải:

a) √6-2√5 + √6+2√5

=√(1-√5)²+√(1+√5)²

=√5 - 1 + 1 + √5

= 2√5

b) √3+2√2 - √6-4√2

= √(1+√2)² - √(2-√2)²

= 1+√2 - (2-√2)

= 1+√2 - 2 + √2

= -1+2√2

c) √6+2√5-√13+4√3

= √6+2√5-(1+2√3)

= √6+2√5-1-2√3

= √6+2√(1-√3)²

= √6+2(√3 - 1)

= √6+2√3 - 2

= √4+2√3

= √(1+√3)²

= 1+√3

d) √4+√8 . √2+√2+√2 . √2-√2+√2

= √(4+√8).(2+√2+√2).(2-√2+√2)

= √(4+2√2).(2+√2+√2).(2-√2+√2)

= (4+2√2).[4-(2+√2)]

= √(4+2√2).(4-2-√2)

= √(4+2√2).(2-√2)

= √8-4√2+4√2-4

= √4

= 2

Bn tham khảo nhé!!!!!!!!!!!!!!!!!!!

Xin ctlhn ạ!!!!!!!!!!!!!!!!!!!!!!