`A=sqrt{3-2sqrt2}-sqrt{3+2sqrt2}`

`=sqrt{2-2sqrt2+1}-sqrt{2+2sqrt2+1}`

`=sqrt{(sqrt2-1)^2}-sqrt{(sqrt2+1)^2}`

`=|sqrt2-1|-|\sqrt2+1|`

`=sqrt2-1-sqrt2-1=-2`

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

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

=-2

a: \(=\sqrt[3]{\dfrac{5}{625}}=\sqrt[3]{\dfrac{1}{125}}=\dfrac{1}{5}\)

b: \(=\sqrt[5]{\left(-\sqrt{5}\right)^5}=-\sqrt{5}\)

Bài 1: 

a) \(\sqrt{72}:\sqrt{8}=\sqrt{72:8}=3\)

b) \(\left(\sqrt{28}-\sqrt{7}+\sqrt{112}\right):\sqrt{7}=5\sqrt{7}:\sqrt{7}=5\)

Bài 2: 

a) \(\sqrt{\dfrac{49}{8}}:\sqrt{3\dfrac{1}{8}}=\sqrt{\dfrac{49}{8}:\dfrac{25}{8}}=\sqrt{\dfrac{49}{25}}=\dfrac{7}{5}\)

b) \(\sqrt{54x}:\sqrt{6x}=\sqrt{54x:6x}=\sqrt{9}=3\)

c) \(\sqrt{\dfrac{1}{125}}\cdot\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}\)

\(=\dfrac{\sqrt{5}}{25}\cdot\dfrac{4\sqrt{2}}{\sqrt{35}}:\dfrac{2\sqrt{14}}{15}\)

\(=\dfrac{\sqrt{5}\cdot4\sqrt{2}\cdot15}{25\cdot\sqrt{35}\cdot\sqrt{14}\cdot2}\)

\(=\dfrac{6}{35}\)

a: \(\sqrt[3]{-125}=\sqrt[3]{\left(-5\right)^3}=-5\)

b: \(\sqrt[4]{\dfrac{1}{81}}=\sqrt[4]{\left(\dfrac{1}{3}\right)^4}=\dfrac{1}{3}\)

a) \(\sqrt[3]{-125}=\sqrt[3]{\left(-5\right)^3}=-5\)

b) \(\sqrt[4]{\dfrac{1}{81}}=\sqrt[4]{\left(\dfrac{1}{3}\right)^4}=\dfrac{1}{3}\)

\(A=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{99}+\sqrt{100}\)

=10-1

=9

a) Ta có: \(-3\sqrt{16}\cdot\sqrt{90}\)

\(=-3\cdot4\cdot3\sqrt{10}\)

\(=-36\sqrt{10}\)

b) Ta có: \(3\sqrt{\dfrac{4}{3}}-3\sqrt{48}+5\sqrt{75}\)

\(=3\cdot\dfrac{2}{\sqrt{3}}-3\cdot4\sqrt{3}+5\cdot5\sqrt{3}\)

\(=2\sqrt{3}-12\sqrt{3}+25\sqrt{3}\)

\(=15\sqrt{3}\)

c) Ta có: \(4\sqrt[3]{27}-\sqrt[3]{64}-2\sqrt[3]{8}\)

\(=4\cdot3-4-2\cdot2\)

\(=12-4-4=4\)

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

b)\(\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+\dfrac{1}{6}=\dfrac{\sqrt{6}}{3}-\sqrt{2^2.6}+\dfrac{2\sqrt{24}}{8}+\dfrac{1}{6}\)

\(=\dfrac{\sqrt{6}}{3}-2\sqrt{6}+\dfrac{\sqrt{2^2.6}}{4}+\dfrac{1}{6}=\dfrac{-5\sqrt{6}}{3}+\dfrac{2\sqrt{6}}{4}+\dfrac{1}{6}\)

\(=\dfrac{-7\sqrt{6}}{6}+\dfrac{1}{6}\)

\(A=\sqrt{243}-\sqrt{27}+\sqrt{3}-\sqrt{48}\\ =\sqrt{81\cdot3}-\sqrt{9\cdot3}+\sqrt{3}-\sqrt{16\cdot3}\\ =9\sqrt{3}-3\sqrt{3}+\sqrt{3}-4\sqrt{4}\\ =\left(9-3+1-4\right)\sqrt{3}\\ =3\sqrt{3}\)

\(B=\dfrac{5+\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{3}+\sqrt{5}\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\sqrt{5}-\sqrt{3}\\ =\sqrt{5}+1+\sqrt{3}-\sqrt{5}-\sqrt{3}\\ =1\)

b: \(=16-2\cdot4\cdot2\sqrt{5}+20-9-4\sqrt{5}\)

=27-20căn 5

a: 2-4căn 3<0

nên biểu thức ko có giá trị

\(b,\left(4-2\sqrt{5}\right)^2-\left(\sqrt{5}+2\right)^2\\ =\left[\left(4-2\sqrt{5}\right)-\left(\sqrt{5}+2\right)\right].\left[\left(4-2\sqrt{5}\right)+\left(\sqrt{5}+2\right)\right]=\left(2-3\sqrt{5}\right)\left(6-\sqrt{5}\right)\)

`b)(4-2\sqrt5)^2-(\sqrt5+2)^2`

`=(4-2\sqrt5-\sqrt5-2)(4-2\sqrt5+\sqrt5+2)`

`=(2-3\sqrt5)(6-\sqrt5)`

$---------$

Áp dụng HĐT :

`a^2-b^2=(a-b)(a+b)`

A: \(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\)

\(=3-3\cdot\dfrac{5\sqrt{2}}{3}+6\sqrt{2}-3\)

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

b: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\cdot\sqrt{\dfrac{16}{3}}\)

\(=\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4}{\sqrt{3}}\)

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

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

\(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\\ =3-3\cdot\dfrac{1}{3}\sqrt{25\cdot2}+3\sqrt{4\cdot2}-3\\ =3-1\cdot5\sqrt{2}+3\cdot2\sqrt{2}-3\\ =3-5\sqrt{2}+6\sqrt{2}-3\\ =\sqrt{2}\)

\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\sqrt{\dfrac{16}{3}}\\ =\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4\sqrt{3}}{3}\\ =2-\sqrt{3}+\sqrt{3}+1-8\sqrt{3}\\ =3-8\sqrt{3}\)