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

\(=\sqrt{6}\left(\sqrt{3}+1\right)\times3\left(2-\sqrt{3}\right)\)

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

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

\(=\dfrac{3\sqrt{6}}{2}\left(3-1\right)\left(\sqrt{3}-1\right)\)

\(=3\sqrt{6}\left(\sqrt{3}-1\right)\)

https://hoc24.vn/hoi-dap/question/405366.html

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

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

\(=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)

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

= 5 - 3

= 2

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

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

câu b hình như sai đề

\(C=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)=\sqrt{3-\sqrt{5}}\sqrt{2}\left(\sqrt{5}-1\right)\)

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

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

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

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

\(A=\left(\sqrt{5}-\sqrt{2}\right)^2+2\sqrt{10}=\left(\sqrt{5.1}-\sqrt{2.1}\right)^2+2\sqrt{10}=7-2\sqrt{10}+2\sqrt{10}=7\)        

Câu B hình như có cái gì đó không ổn          

\(C=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)=\sqrt{3-\sqrt{5}}\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2\left(3-\sqrt{5}\right)}\left(\sqrt{5-1}\right)=\sqrt{6-2\sqrt{5}}\left(-1+\sqrt{5}\right)=6-2\sqrt{5}\)

\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)

\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)

\(d,\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)\\ =\left(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right)\sqrt{2}\left(\sqrt{5}-1\right)\\ =\left(2\sqrt{4+\sqrt{5}-1}\right)\sqrt{2}\left(\sqrt{5}-1\right)\\ =\sqrt{24+8\sqrt{5}}\left(\sqrt{5}-1\right)\\ =\sqrt{\left(2\sqrt{5}+2\right)^2}\left(\sqrt{5}-1\right)\\ =2\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\\ =2\left(5-1\right)\\ =8\)

Xem lại câu c) và d) 

b: =căn 10-3+4-căn 10=1

a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)

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

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

b/ \(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)

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

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

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

\(=2\left(3-4\right)=-2\)

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

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

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

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

d/ \(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

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

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(16-15\right)=2\)