Câu 1: 

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

Câu 2: 

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=2\sqrt{3}\\2x-3=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{3}+3}{2}\\x=\dfrac{-2\sqrt{3}+3}{2}\end{matrix}\right.\)

`1)A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}`

   `A=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}`

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

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

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

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`2)\sqrt{4x^2-12x+9}=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}`  

`<=>\sqrt{4x^2-12x+9}=2\sqrt{2}` (Như câu `1`)

`<=>4x^2-12x+9=8`

`<=>4x^2-12x+1=0`

Ptr có:`\Delta'=(-6)^2-4=32 > 0`

`=>` Ptr có `2` nghiệm pb

`x_1=[-b+\sqrt{\Delta'}]/a=[-(-6)+\sqrt{32}]/4=[3+2\sqrt{2}]/2`

`x_2=[-b-\sqrt{\Delta'}]/a=[-(-6)-\sqrt{32}]/4=[3-2\sqrt{2}]/2`

Vậy `S={[3+-2\sqrt{2}]/2}`

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

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

\(=\dfrac{20-6}{2}=7\)

a: \(E=1+1=2\)

b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)

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

d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)

Đặt \(x=\sqrt{\dfrac{5+2\sqrt{6}}{5-\sqrt{6}}}+\sqrt{\dfrac{5-2\sqrt{6}}{5+\sqrt{6}}}>0\)

\(x^2=\dfrac{5+2\sqrt{6}}{5-\sqrt{6}}+\dfrac{5-2\sqrt{6}}{5+\sqrt{6}}+2\sqrt{\dfrac{25-24}{25-6}}=\dfrac{74}{19}+\dfrac{2\sqrt{19}}{19}\)

\(\Rightarrow x^2=\dfrac{74+2\sqrt{19}}{19}\Rightarrow x=\sqrt{\dfrac{74+2\sqrt{19}}{19}}\)

Ko thể rút gọn thêm nữa (có thể trục căn thức ở mẫu)

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.

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

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

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

chịu 😅

, \(A=\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)

\(=\frac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+\frac{6\left(\sqrt{7}+2\right)}{3}-\frac{5\left(4-\sqrt{7}\right)}{9}\)

\(=\frac{-16+4\sqrt{7}}{4}+\frac{18\sqrt{7}+36-20+5\sqrt{7}}{9}=-4+\sqrt{7}+\frac{23\sqrt{7}+16}{9}\)

b,\(B=\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}=\frac{2\left(\sqrt{6}+2\right)+2\left(\sqrt{6}-2\right)}{2}+\frac{5\sqrt{6}}{6}\)

\(=\frac{12\sqrt{6}+5\sqrt{6}}{6}=\frac{17\sqrt{6}}{6}\)

a,32 căn 7 -20/9

b, 17 căn 6 / 6

a: A=2-căn 3+căn 3=2

b: B=căn 5+1-căn 5+1=2