`\sqrt{(3-2\sqrt{2})^2}+\sqrt{(3+2\sqrt{2})^2}`

`=|3-2\sqrt{2}|+3+2\sqrt{2}`

`=3-2\sqrt{2}+3+2\sqrt{2}=6`

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

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

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

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

\(A=\left|2-\sqrt{7}\right|+7-2\sqrt{7}+1\)

\(=\sqrt{7}-2+8-2\sqrt{7}\) \(=6-\sqrt{7}\)

\(B=3\cdot1,5-4\cdot\left|3-\sqrt{2}\right|\) \(=4,5-4\left(3-\sqrt{2}\right)\)

\(=4,5-12+4\sqrt{2}\) \(=4\sqrt{2}-7,5\) 

Ta có: \(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)

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

\(=6-\sqrt{7}\)

Ta xét:

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

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

\(a=6-4\sqrt{2}-3\)

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

và

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

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

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

\(b=\sqrt{3+2\sqrt{2}}\cdot\sqrt{9-8}\)

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

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

\(=2\left|3-\sqrt{2}\right|+\sqrt{18}-5.1=6-2\sqrt{2}+3\sqrt{2}-5\)

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

\(1,=\left|1-\sqrt{2}\right|+\left|\sqrt{2}+3\right|\\ =1-\sqrt{2}+3+\sqrt{2}\\ =4\\ 2,=\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}-2+\sqrt{3}-1\\ =2\sqrt{3}-3\\ 3,=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{5}-3+\sqrt{5}-2\\ =2\sqrt{5}-5\\ 4,=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =3+\sqrt{3}\\ 5,=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\\ =2-\sqrt{3}-\left(2+\sqrt{3}\right)\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)

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

Lời giải:

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

$=3\sqrt{2}$

b. $=|\sqrt{7}-2\sqrt{2}|-|\sqrt{7}+2\sqrt{2}|$

$=(2\sqrt{2}-\sqrt{7})-(\sqrt{7}+2\sqrt{2})$

$=-2\sqrt{7}$

c.

$=|3+\sqrt{5}|+|3-\sqrt{5}|=(3+\sqrt{5})+(3-\sqrt{5})=6$

d.

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

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

\(A=-7\)

Ta có: \(A=\left(\sqrt{a}+2\right)\left(\sqrt{a}-3\right)-\left(\sqrt{a}+1\right)^2+\sqrt{9a}\)

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

\(=-7\)