\(\sqrt{11-2\sqrt{18}}=3-\sqrt{2}\)

=> a=3; b=-1

\(\sqrt{11+2\sqrt{18}}=\sqrt{9+2.3.\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\Rightarrow ab=3\)

\(\sqrt{11-2\sqrt{18}}=\sqrt{11-2\sqrt{9.2}}=\sqrt{\left(\sqrt{2}\right)^2-2.3\sqrt{2}+9}\) =\(\sqrt{\left(3-\sqrt{2}\right)^2}\)=  \(3-\sqrt{2}\)

=> a=3, b=-1   => ab =-3

đề hỏi j

\(A=9\sqrt{2}+7\sqrt{2}-12\sqrt{2}=4\sqrt{2}\)
\(B=\sqrt{2}-1-\sqrt{\left(3+\sqrt{2}\right)^2}=\sqrt{2}-1-3-\sqrt{2=-4}\)

\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)

a)
\(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{3}+\sqrt{2}-\sqrt{2}+1=\sqrt{3}+1\)
b)
\(\sqrt{\left(\sqrt{9}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{16}+\sqrt{2}\right)^2}=\sqrt{9}+\sqrt{2}-\sqrt{16}-\sqrt{2}=3-4=-1\)
c)
\(=\sqrt{2\left(2-\sqrt{3}\right)}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)

\(\sqrt{11-2\sqrt{18}}=\sqrt{2-2.3.\sqrt{2}+9}=3-\sqrt{2}\Rightarrow ab=-3\)

a: \(\dfrac{1}{3}\cdot\sqrt{18}-\sqrt{192}-\dfrac{\sqrt{33}}{\sqrt{11}}+3\cdot\sqrt{5\dfrac{1}{3}}\)

\(=\dfrac{1}{3}\cdot3\sqrt{2}-8\sqrt{3}-\sqrt{3}+3\cdot\dfrac{4}{\sqrt{3}}\)

\(=\sqrt{2}-7\sqrt{3}+4\sqrt{3}\)

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

b: Ta có: \(\sqrt{\left(2\sqrt{3}-5\right)^2}-2\cdot\sqrt{7+4\sqrt{3}}\)

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

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

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