a)

\(\left(3-\sqrt{15}\right)\sqrt{4+\sqrt{15}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{5+2\sqrt{15}+3}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}\\ =\left(\sqrt{9}-\sqrt{15}\right)\cdot\dfrac{\left|\sqrt{5}+\sqrt{3}\right|}{\sqrt{2}}\)

\(=\sqrt{3}\left(\sqrt{3}-\sqrt{5}\right)\cdot\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\) (vì \(\sqrt{5}+\sqrt{3}>0\))

\(=\sqrt{3}\cdot\dfrac{3-5}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-2}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-\sqrt{4}}{\sqrt{2}}\\ =-\sqrt{6}\)

b)

\(\sqrt{29-12\sqrt{5}}-\sqrt{24-8\sqrt{5}}\\ =\sqrt{20-2\cdot3\cdot2\sqrt{5}+9}-\sqrt{20-2\cdot2\cdot2\sqrt{5}+4}\\ =\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(2\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}-3\right|-\left|2\sqrt{5}-2\right|\)

\(=2\sqrt{5}-3-\left(2\sqrt{5}-2\right)\) (vì \(2\sqrt{5}-3>0;2\sqrt{5}-2>0\))

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

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

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

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

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

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

1) \(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)

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

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

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

\(=4\)

2) \(\sqrt{12-6\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)

\(=\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{3^2+2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

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

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

\(=6\)

\(\Rightarrow P=4-2+2=4\)

P=\(\sqrt{16}-\sqrt[3]{8}+\dfrac{\sqrt{12}}{\sqrt{3}}\)⇔ P= 4-2+\(\sqrt{\dfrac{12}{3}}\)

➜ P= 4-2+2 = 4

\(P=\sqrt{16}-\sqrt[3]{8}+\dfrac{\sqrt{12}}{\sqrt{3}}\)

\(=4-2+2\)

=4

\(\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)

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

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

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

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

\(=5\)

\(A=\sqrt{\left(9\sqrt{2}+2\sqrt{3}\right)^2}-\sqrt{\left(9\sqrt{2}-\sqrt{3}\right)^2}\)

\(=\left|9\sqrt{2}+2\sqrt{3}\right|-\left|9\sqrt{2}-\sqrt{3}\right|\)

\(=9\sqrt{2}+2\sqrt{3}-9\sqrt{2}+\sqrt{3}=3\sqrt{3}\)

Kiểm tra lại đề bài câu B, chỗ \(\sqrt{2+\sqrt{2+2}}\)

Nếu câu B sửa đề thành:

\(B=\sqrt{2+\sqrt{2}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)

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

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

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

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

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