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

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

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

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

\(P=\dfrac{a+2\sqrt{a}}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\\ =\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}=\sqrt{a}-\left(\sqrt{a}+2\right)=-2\)

Ta có: \(P=\dfrac{a+2\sqrt{a}}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\)

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

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

Lời giải:
\(Q=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

\(=\frac{(1+\sqrt{2})(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

A=\(\sqrt{\left(\sqrt{7}-2\right)^2}\)+\(\frac{25\sqrt{7}-63}{3\sqrt{7}-7}\)=\(\frac{12\sqrt{7}-28}{3\sqrt{7}-7}\)=4

Thiếu đề

A=−√32

A=−0.86602540…

\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)

\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)

=> \(A=\sqrt{5}\)