\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-2.3.\sqrt{6}}+\sqrt{33-2.6\sqrt{6}}\)

\(=\sqrt{3^2-2.3.\sqrt{6}+\sqrt{6^2}}+\sqrt{24-2.2\sqrt{6}.3+9}\)

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

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

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

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

Ko vt lại đề nha bn:

\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-2.3.\sqrt{6}}+\sqrt{33-2.6\sqrt{6}}\)

\(=\sqrt{3^2-2.3.\sqrt{6}+\sqrt{6^2}}+\sqrt{24-2.2\sqrt{6}.3+9}\)

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

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

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

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

Rất vui vì giúp đc bn !!!

 √(15 - 6√6)

=√(9-2.3.√6+6)

=√(3-√6)^2=3-√6

Chúc các bn hok tốt

Bn ko nên tự trả lời câu hỏi của mik như vậy

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

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

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

b) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

c) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)

= \(\dfrac{15\sqrt{200}}{\sqrt{10}}-\dfrac{3\sqrt{450}}{\sqrt{10}}+\dfrac{2\sqrt{50}}{\sqrt{10}}\)

= \(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\)

= \(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\) = \(23\sqrt{5}\)

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

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

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

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

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

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

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

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

\(A=-3\)

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

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

\(B=1\)

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

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

\(=-3\)

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

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

\(=1\)

mấy giờ nộp để đây làm cho ! Chờ đây ăn cơm xong đã!

\(\sqrt{3^2-2.3.\sqrt{6}+6}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2.\sqrt{6}\right)^2}\)

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