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

\(=16-3+3-16=0\)

a: \(x=4+\sqrt{3}+4-\sqrt{3}=8\)

Khi x=8 thì \(A=\dfrac{2-5\cdot2\sqrt{2}}{2\sqrt{2}+1}=\dfrac{2-10\sqrt{2}}{2\sqrt{2}+1}=-6+2\sqrt{2}\)

\(A=\left(4+\sqrt{3}\right)\sqrt{19-8\sqrt{3}}\)

\(A=\left(4+\sqrt{3}\right)\sqrt{4^2-2\cdot4\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

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

\(A=4^2-3\)

\(A=13\)

\(B=\dfrac{3}{4+\sqrt{13}}+\dfrac{\sqrt{52}}{2}-3\)

\(B=\dfrac{3\left(4-\sqrt{13}\right)}{\left(4-\sqrt{13}\right)\left(4+\sqrt{13}\right)}+\dfrac{2\sqrt{13}}{2}-3\)

\(B=\dfrac{3\left(4-\sqrt{13}\right)}{16-13}+\sqrt{13}-3\)

\(B=4-\sqrt{13}+\sqrt{13}-3\)

\(B=4-3\)

\(B=1\)

`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}`

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

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

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

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

a: Ta có: \(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)

\(=5-\sqrt{19}-\sqrt{19}+4\)

\(=9-2\sqrt{19}\)

b: Ta có: \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}\)

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

=0

c.

Căn bậc 2 không xác định do $2-\sqrt{5}< 0$

d.

\(=\sqrt{(3+\sqrt{3})^2}(3+\sqrt{3})=|3+\sqrt{3}|(3+\sqrt{3})=(3+\sqrt{3})^2=12+6\sqrt{3}\)

e.

\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)

a) \(A=\sqrt{19+8\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)

\(A=\sqrt{16+8\sqrt{3}+3}-\sqrt{3+2\sqrt{3}+1}\)

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

\(A=4+\sqrt{3}-\sqrt{3}-1=3\)

b) \(B=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(B=\sqrt{25+10\sqrt{2}+2}-\sqrt{16+8\sqrt{2}+2}\)

\(A=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)

\(A=5+\sqrt{2}-4-\sqrt{2}=1\)

\(A=\sqrt{19+8\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{3+8\sqrt{3}+16}-\sqrt{3+2\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot4+4^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}+1^2}\)

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

\(=\left|\sqrt{3}+4\right|-\left|\sqrt{3}+1\right|\)

\(=\sqrt{3}+4-\left(\sqrt{3}+1\right)\)

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

\(B=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=\sqrt{2+10\sqrt{2}+25}-\sqrt{2+8\sqrt{2}+16}\)

\(=\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot5+5^2}-\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot4+4^2}\)

\(=\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{\left(\sqrt{2}+4\right)^2}\)

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

\(=\sqrt{2}+5-\left(\sqrt{2}+4\right)\)

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

\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)

\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)

=> \(A=\sqrt{5}\)

b, t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).(\(\sqrt{10}\)-\(\sqrt{2}\))

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

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

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

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

t = 15 + \(5\sqrt{5}\) \(-6\sqrt{5}\)-10+1+\(\sqrt{5}\)

t = 6