a) \(\sqrt{19-6\sqrt{2}}=3\sqrt{2}-1\)

b) \(\sqrt{11-6\sqrt{2}}=3-\sqrt{2}\)

d) \(\sqrt{21+12\sqrt{3}}=2\sqrt{3}+3\)

e) \(\sqrt{57-40\sqrt{2}}=4\sqrt{2}-5\)

\(a.\sqrt{19-6\sqrt{2}}=\sqrt{18-2.3\sqrt{2}+1}=3\sqrt{2}-1\)

\(b.\sqrt{21+12\sqrt{3}}=\sqrt{12+2.2\sqrt{3}.3+9}=2\sqrt{3}+3\)

\(c.\sqrt{57-40\sqrt{2}}=\sqrt{32-2.4\sqrt{2}.5+25}=4\sqrt{2}-5\)

\(d.\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{3-2\sqrt{3}.\sqrt{2}+2}.\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\) \(e.\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\) \(g.\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}=\sqrt{4-2.2\sqrt{3}+3}-\sqrt{4+2.2\sqrt{3}+3}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)

a)

=\(\sqrt{18-2.3\sqrt{2}.1+1}\)

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

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

b)

=\(\sqrt{12+2.2\sqrt{3}.3+9}\)

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

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

c)

=\(\sqrt{25-2.5.4\sqrt{2}+32}\)

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

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

d)

\(=\sqrt{\left(3-2.\sqrt{3}.\sqrt{2}+2\right)\left(3-2\sqrt{3}+1\right)}\\ =\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2\left(\sqrt{3}-1\right)^2}\\ =\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\\ =3-\sqrt{3}-\sqrt{6}+\sqrt{2}\)

e)

\(=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}\\ =\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\\ =3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\\ =6\sqrt{2}\)

g)

\(=\sqrt{4-2.2.\sqrt{3}+3}-\sqrt{4+2.2.\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)

b, \(\sqrt{21+12\sqrt{3}}=\sqrt{21+2.3.2.\sqrt{3}}=\sqrt{9+2.3.\sqrt{12}+12}\)

\(=\sqrt{\left(3+\sqrt{12}\right)^2}=3+\sqrt{12}\)

\(c,\sqrt{57-40\sqrt{2}}=\sqrt{57-2.4.5.\sqrt{2}}=\sqrt{25-2.5.\sqrt{32}}\)

\(=\sqrt{\left(5-\sqrt{32}\right)^2}=\left|5-\sqrt{32}\right|=5-\sqrt{32}\)

\(d,\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{\left(3-2.\sqrt{2}.\sqrt{3}+2\right)\left(3-2\sqrt{3}+1\right)}\) \(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2\left(\sqrt{3}-1\right)^2}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\)

\(e,A=\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}\)

Ta có :

\(21+6\sqrt{6}=\dfrac{42+12\sqrt{6}}{2}=\dfrac{36+2.6.\sqrt{6}+6}{2}=\left(\dfrac{6+\sqrt{6}}{\sqrt{2}}\right)^2\) Tương tự : \(21-6\sqrt{6}=\left(\dfrac{6-\sqrt{6}}{\sqrt{2}}\right)^2\)

Do đó :

\(A=\sqrt{\left(\dfrac{6+\sqrt{6}}{\sqrt{2}}\right)^2}+\sqrt{\left(\dfrac{6-\sqrt{6}}{\sqrt{2}}\right)^2}=\dfrac{6+\sqrt{6}}{\sqrt{2}}+\dfrac{6-\sqrt{6}}{\sqrt{2}}=\dfrac{6+\sqrt{6}+6-\sqrt{6}}{\sqrt{2}}\)\(=\dfrac{12}{\sqrt{2}}=\dfrac{12\sqrt{2}}{2}=6\sqrt{2}\)

Phần g làm tương tự như phần e nha bạn :>

Chúc bạn học tốt :>

\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)

ai lam ho voi

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

Giờ mình chỉ giải đc câu a thôi để hồi nao mình rảnh giải típ cho

ok ban

a) \(E=\sqrt{\left|12\sqrt{5}-29\right|}-\sqrt{12\sqrt{5}+29}\)

\(\Leftrightarrow E^2=\left|12\sqrt{5}-29\right|-12\sqrt{5}-29\)

\(\Leftrightarrow E^2=29-12\sqrt{5}-12\sqrt{5}-29\)

\(\Leftrightarrow E^2=-24\sqrt{5}\)

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

b) Đặt \(F=\sqrt{\left|40\sqrt{2}-57\right|}-\sqrt{40\sqrt{2}+57}\)

\(\Leftrightarrow F^2=\left|40\sqrt{2}-57\right|-40\sqrt{2}-57\)

\(\Leftrightarrow F^2=57-40\sqrt{2}-40\sqrt{2}-57\)

\(\Leftrightarrow F^2=-80\sqrt{2}\)

\(\Leftrightarrow F=-4\sqrt{5\sqrt{2}}\)

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.

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

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

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