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

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

tích mình đi

ai tích mình 

mình tích lại 

thanks

con cacacacacacacacacacacacacacacacacacca

@@22@22@22@@222@@2@@2@@@2@2

bạn kiểm tra lại đề bài cấu (c)

\(\sqrt{\frac{6-2\sqrt{5}}{2}}.\sqrt{2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)\(=\frac{\left(\sqrt{5}-1\right)}{\sqrt{2}}.\sqrt{2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=2\left(9-5\right)=2.4=8\)

Bài 1:

a) Ta có: \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

\(=\sqrt{45-2\cdot\sqrt{45}\cdot1+1}-\sqrt{9-2\cdot\sqrt{9}\cdot\sqrt{20}+20}\)

\(=\sqrt{\left(\sqrt{45}-1\right)^2}-\sqrt{\left(3-\sqrt{20}\right)^2}\)

\(=\left|\sqrt{45}-1\right|-\left|3-\sqrt{20}\right|\)

\(=\sqrt{45}-1-3+\sqrt{20}\)

\(=\sqrt{45}+\sqrt{20}-4\)

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

b) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{8}+8}-\sqrt{45+2\cdot\sqrt{45}\cdot\sqrt{8}+8}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{8}\right)^2}-\sqrt{\left(\sqrt{45}+\sqrt{8}\right)^2}\)

\(=\left|\sqrt{5}-\sqrt{8}\right|-\left|\sqrt{45}+\sqrt{8}\right|\)

\(=\sqrt{8}-\sqrt{5}-\sqrt{45}-\sqrt{8}\)

\(=-\sqrt{5}-\sqrt{45}=-\sqrt{5}\left(1+\sqrt{9}\right)=-4\sqrt{5}\)

c) Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{7+4\sqrt{3}}\)

\(=\left(3-\sqrt{2}\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\)

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

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

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

d) Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10+2\sqrt{21}}\)

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

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

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

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

a) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\) = \(6+\sqrt{15}-2\sqrt{15}\)

= \(6-\sqrt{15}\)

b) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\) = \(5\sqrt{10}+10-5\sqrt{10}\) = \(10\)

c) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\) = \(14-2\sqrt{21}-7+2\sqrt{21}\)

= \(7\)

d) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

= \(33-3\sqrt{22}-11+3\sqrt{22}\) = \(22\)

a)(2√3+√5)√3-√60
=6+√15-2√15
=6-√15

b)(5√2+2√5)√5-√250
=5√10+10-5√10
=10

c)(√28-√12-√7)√7+2√21
=14-2√21-7+2√21
=7

d)(√99-√18-√11)√11+3√22
=33-3√22-11+3√22
=22