b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)

\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)

\(=4\left(7+3\sqrt{5}\right)\)

\(=28+12\sqrt{5}\)

Lời giải:

a. 

$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$

$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$

$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$

$=2\sqrt{5}-5\sqrt{10}$

$\Rightarrow A=\sqrt{10}-5\sqrt{5}$

b.

$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$

$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$

$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$

$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$

$\Rightarrow B=28+12\sqrt{5}$

c.

$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$

$=(7-5)(6-\sqrt{35})$

$=2(6-\sqrt{35})=12-2\sqrt{35}$

`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`

`=`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`

`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`

`=2(\sqrt5-1)sqrt{6+2\sqrt5}`

`=2(\sqrt5-1)(\sqrt5+1)`

`=2(5-1)`

`=8`

`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`

`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`

`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`

`=2(\sqrt5-1)sqrt{6+2\sqrt5}`

`=2(\sqrt5-1)(\sqrt5+1)`

`=2(5-1)`

`=8`

`(4\sqrt2+\sqrt{30})(\sqrt5-\sqrt3)\sqrt{4-\sqrt{15}}`

`=\sqrt2(4+\sqrt{15})(\sqrt5-\sqrt3)\sqrt{4-\sqrt{15}}`

`=(4+\sqrt{15})(\sqrt5-\sqrt3)\sqrt{8-2\sqrt{15}}`

`=(4+\sqrt{15})(\sqrt5-\sqrt3)(\sqrt5-\sqrt3)`

`=(4+\sqrt{15})(8-2\sqrt{15})`

`=2(4+\sqrt{15})(4-\sqrt{15})`

`=2(16-15)`

`=2`

a) \(2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{6-2\sqrt{5}}}\)

\(=2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{5}-1}\)

\(=\dfrac{2\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{5}+1\right)}{\sqrt{2}}\)

\(=2\cdot4=8\)

Câu 1,2 bạn đã đăng và có lời giải rồi

Câu 3:

\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)

ai nhanh nhat mk se k (neu dung). mk cần gấp

tích mình đi

ai tích mình 

mình tích lại 

thanks

bn trả lời đúng đi rồi mk k. thank

a: \(A=\dfrac{2\sqrt{2}\left(\sqrt{3}+1\right)}{3\cdot\sqrt{2+\sqrt{3}}}=\dfrac{4\left(\sqrt{3}+1\right)}{3\cdot\sqrt{4+2\sqrt{3}}}\)

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

b: \(B=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\left|3\sqrt{5}-3\right|\)

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

\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)

\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)

\(=\sqrt{29}\)

câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :

\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)

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

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

\(=3-\sqrt{6}+2\sqrt{6}-3\)   ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )

\(=\sqrt{6}\)

A = \(\dfrac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)

A = \(\dfrac{\sqrt{3}+\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)

A = \(\dfrac{\sqrt{3}+\sqrt{2}+3-\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{5}+1-\left(\sqrt{5}+\sqrt{2}\right)}\)

A = \(\dfrac{\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}-\sqrt{2}}\) = \(\dfrac{3}{1}\) = \(3\)

C = \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

C = \(\left(4+\sqrt{15}\right).\left(\sqrt{40-10\sqrt{15}}-\sqrt{24-6\sqrt{15}}\right)\)

C = \(\left(4+\sqrt{15}\right)\left(\sqrt{\left(5-\sqrt{15}\right)^2}-\sqrt{\left(\sqrt{15}-3\right)^2}\right)\)

C = \(\left(4+\sqrt{15}\right)\left(5-\sqrt{15}-\left(\sqrt{15}-3\right)\right)\)

C = \(\left(4+\sqrt{15}\right)\left(5-\sqrt{15}-\sqrt{15}+3\right)\)

C = \(\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

C = \(32-8\sqrt{15}+8\sqrt{15}-30=2\)

D = \(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)

D = \(\left(\sqrt{30-10\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)\left(3+\sqrt{5}\right)\)

D = \(\left(\sqrt{\left(5-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\right)\left(3+\sqrt{5}\right)\)

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

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

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

D = \(18+6\sqrt{5}-6\sqrt{5}-10=8\)

E = \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{5}}\)

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

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