a)\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

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

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

\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

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

\(=5-\sqrt{15}+\sqrt{15}-3\)

\(=2\)

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

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

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

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

\(=6\sqrt{10}+2\sqrt{50}-6\sqrt{2}-2\sqrt{10}\)

\(=6\sqrt{10}+10\sqrt{2}-6\sqrt{2}-2\sqrt{10}\)

\(=4\sqrt{10}+4\sqrt{2}\)

c) \(\left(\sqrt{7}+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)

\(=\left(\sqrt{7}+\sqrt{14}\right)\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}\)

\(=\left(\sqrt{7}+\sqrt{14}\right)\left(\sqrt{7}-\sqrt{2}\right)\)

\(=7\sqrt{7}-7\sqrt{2}+\sqrt{98}-\sqrt{28}\)

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

\(=5\sqrt{7}\)

d) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)

\(=\sqrt{\dfrac{289+42\sqrt{2}}{16}}\)

\(=\dfrac{\sqrt{289+42\sqrt{2}}}{\sqrt{4^2}}\)

\(=\dfrac{\sqrt{\left(1+12\sqrt{2}\right)^2}}{4}\)

\(=\dfrac{1+12\sqrt{2}}{4}\)

e) \(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}\)

\(=\left(\sqrt{21}+\sqrt{7}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}\)

\(=\left(\sqrt{21}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\sqrt{147}-\sqrt{63}+7-\sqrt{21}\)

\(=7\sqrt{3}-\sqrt{63}+7-\sqrt{21}\)

f) bạn xem đề lại nhé

g) \(\sqrt{2}\sqrt{8+3\sqrt{7}}\)
\(=\sqrt{2\left(8+3\sqrt{7}\right)}\)

\(=\sqrt{16+6\sqrt{7}}\)

\(=\sqrt{\left(3+\sqrt{7}\right)^2}\)

\(=3+\sqrt{7}\)

g)

\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)

\(\sqrt{8-2\sqrt{12}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}=\left|\sqrt{6}-\sqrt{2}\right|=\sqrt{6}-\sqrt{2}\)

\(\sqrt{21+6\sqrt{6}}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\left|3\sqrt{2}-\sqrt{3}\right|=3\sqrt{2}-\sqrt{3}\)

\(\sqrt{15-6\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|=3-\sqrt{6}\)

\(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)

\(\sqrt{41+12\sqrt{5}}=\sqrt{\left(6+\sqrt{5}\right)^2}=6+\sqrt{5}\)

Bài 1:

a) Ta có: \(\sqrt{\left(23-15\sqrt{3}\right)^2}\)

\(=\left|23-15\sqrt{3}\right|\)

\(=\left|\sqrt{529}-\sqrt{675}\right|\)

\(=\sqrt{675}-\sqrt{529}\)

\(=15\sqrt{3}-23\)

b) Ta có: \(\sqrt{\left(2-2\sqrt{3}\right)^2}\)

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

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

c) Ta có: \(\sqrt{\left(15-4\sqrt{3}\right)^2}\)

\(=\left|15-4\sqrt{3}\right|\)

\(=15-4\sqrt{3}\)

d) Ta có: \(\sqrt{\left(16-6\sqrt{7}\right)^2}\)

\(=\left|16-6\sqrt{7}\right|\)

\(=\left|\sqrt{256}-\sqrt{252}\right|\)

\(=16-6\sqrt{7}\)

f) Ta có: \(\sqrt{\left(22-8\sqrt{3}\right)^2}\)

\(=\left|22-8\sqrt{3}\right|\)

\(=\left|\sqrt{484}-\sqrt{192}\right|\)

\(=22-8\sqrt{3}\)

g) Ta có: \(\sqrt{\left(9-4\sqrt{2}\right)^2}\)

\(=\left|9-4\sqrt{2}\right|\)

\(=9-4\sqrt{2}\)

h) Ta có: \(\sqrt{\left(13-4\sqrt{3}\right)^2}\)

\(=\left|13-4\sqrt{3}\right|\)

\(=13-4\sqrt{3}\)

i) Ta có: \(\sqrt{\left(7-3\sqrt{3}\right)^2}\)

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

\(=7-3\sqrt{3}\)

Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)

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

=1

\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)

\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)

`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`

`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`

`=root{3}{4(1-sqrt3)(1+sqrt3)}`

`=root{3}{4(1-3)}=-2`

`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`

`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`

`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`

`=root{3}{9}`

`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`

`=root{3}{(8sqrt5-16)(8sqrt5+16)}`

`=root{3}{320-256}`

`=root{3}{64}=4`

`b)root{3}{7-5sqrt2}-root{6}{8}`

`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`

`=root{3}{(1-sqrt2)^3}-sqrt2`

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

a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)

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

=-4

b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)

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

\(=3\sqrt{3}+1\)

c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)

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

\(=3\sqrt{5}-6\)

d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)

\(=\sqrt{7}-2+4-\sqrt{7}+8\)

=10

a) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\left(\sqrt{15}-\sqrt{6}\right)\left(\sqrt{35}+\sqrt{14}\right)}{21}\)

\(=\dfrac{\sqrt{525}+\sqrt{210}-\sqrt{210}-\sqrt{84}}{21}=\dfrac{5\sqrt{21}-2\sqrt{21}}{21}\)

\(=\dfrac{3\sqrt{21}}{21}=\dfrac{\sqrt{21}}{7}\)

b) \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{10}+\sqrt{15}}{2\sqrt{2}+2\sqrt{3}}\)

\(=\dfrac{\left(\sqrt{10}+\sqrt{15}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{-4}=\dfrac{\left(\sqrt{10}+\sqrt{15}\right)\left(\sqrt{2}-\sqrt{3}\right)}{-2}\)

\(=\dfrac{\left(\sqrt{10}+\sqrt{15}\right)\left(\sqrt{2}-\sqrt{3}\right)}{-2}=\dfrac{\sqrt{20}-\sqrt{30}+\sqrt{30}-\sqrt{45}}{-2}\)

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

c) \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\) có sai k nhỉ

d) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) (tự làm đc kq là \(1+\sqrt{2}\))

e,f) xem lại đề