\(a,A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)

\(=\sqrt{\left(\sqrt{5}^2+2\sqrt{5}+2\sqrt{2}\cdot\sqrt{5}\right)+\sqrt{2}^2+2\sqrt{2}\cdot1+1^2}\)

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

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

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

\(b,B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\frac{3\cdot\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}+\frac{2\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}{\sqrt{6}-2}-\frac{4\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[3\cdot\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}+11\right)\left(\sqrt{6}-11\right)=-115\)

\(\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}\)=\(\sqrt{2+5+1+2\sqrt{2}+2\sqrt{5}+2\sqrt{2}\cdot\sqrt{5}}\)

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

\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)

\(=\sqrt{1+2+5+2\left(\sqrt{2}+\sqrt{5}+\sqrt{10}\right)}=\sqrt{\left(1+\sqrt{2}+\sqrt{5}\right)^2}\)

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

\(A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}=\sqrt{\sqrt{5}^2+\sqrt{2}^2+1^2+2\sqrt{2}.1}+2\sqrt{5}.1+2\sqrt{2}\sqrt{5}}\)

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

\(A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)

\(=\sqrt{\sqrt{5}^5+\sqrt{2}^2+1^2+2\sqrt{2}.1+2\sqrt{2}.\sqrt{5}}\)

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

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

=√√55+√22+12+2√2.1+2√2.√5

=√(√5+√2+1)2

\(A=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+....}}}}>0\)

\(\Rightarrow A^2=6+\sqrt{6+\sqrt{6+\sqrt{6+....}}}\)

\(\Rightarrow A^2=6+A\)\(\Rightarrow A^2-A-6=0\)

\(\Rightarrow\left(A-3\right)\left(A+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}A-3=0\\A+2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}A=3\\A=-3\end{cases}}\Rightarrow A=3>0\) (thỏa)

câu 1 mình làm được rồi! mik cần mọi người help mình câu 2 ! pleaseeeeee.......... T-T

Ta có: \(\dfrac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)

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

\(=\sqrt{2}\)

a: \(=2\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)

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

b: \(=2\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}\)