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

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

SBT nha

B=\(\sqrt{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}\)

B=\(\sqrt{7+5+2+2\sqrt{2.5}+2\sqrt{2.7}+2\sqrt{7.5}}\)

B=\(\left(\sqrt{\sqrt{7}+\sqrt{5}+\sqrt{2}}\right)^2\)

B=\(\sqrt{7}+\sqrt{5}+\sqrt{2}\)

\(P=\sqrt{\left(\sqrt{2}+\sqrt{5}+\sqrt{7}\right)^2}=\left|\sqrt{2}+\sqrt{5}+\sqrt{7}\right|=\sqrt{2}+\sqrt{5}+\sqrt{7}\)

Ta có

\(P=\sqrt{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}\)

\(\Leftrightarrow P=\sqrt{\left(\sqrt{5}+\sqrt{2}+\sqrt{7}\right)^2}\)

\(\Leftrightarrow P=\sqrt{5}+\sqrt{2}+\sqrt{7}\)

Mà \(P=\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{5}+\sqrt{2}+\sqrt{7}\)

Suy ra \(a+b+c=5+2+7=14\)

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

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

=> \(A=\sqrt{5}\)

P=\(\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}\)=\(\sqrt{2+5+7+2\sqrt{5.2}+2\sqrt{2.7}+2\sqrt{3.5}}\)

=\(\sqrt{\left(\sqrt{2}+\sqrt{5}+\sqrt{7}\right)^2}\)=\(\sqrt{2}+\sqrt{5}+\sqrt{7}\)=\(\sqrt{a}+\sqrt{b}+\sqrt{c}\)

Vậy a+b+c=14

14

14

\(\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}\)

\(=\sqrt{2+5+7+2\sqrt{2.5}+2\sqrt{2.7}+2\sqrt{5.7}}\)

\(=\sqrt{\left(\sqrt{2}+\sqrt{5}+\sqrt{7}\right)^2}=\sqrt{2}+\sqrt{5}+\sqrt{7}\)

\(\Rightarrow a+b+c=2+5+7=14\)