\(A=\sqrt{27}-2\sqrt{12}-\sqrt{75}\)

\(A=\sqrt{9.3}-2\sqrt{3.4}-\sqrt{25.3}\)

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

\(A=-6\sqrt{3}\)

\(B=\frac{1}{3+\sqrt{7}}+\frac{1}{3-\sqrt{7}}\)

\(B=\frac{3-\sqrt{7}+3\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)

\(B=\frac{6}{9-7}=3\)

\(A=\sqrt{27}-2\sqrt{12}-\sqrt{75}\)

\(=\sqrt{3^2.3}-2.\sqrt{2^2.3}-\sqrt{5^2.3}\)

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

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

vậy \(A=-6\sqrt{3}\)

\(B=\frac{1}{3+\sqrt{7}}+\frac{1}{3-\sqrt{7}}\)

\(B=\frac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}+\frac{3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}\)

\(B=\frac{3-\sqrt{7}+3+\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)

\(B=\frac{6}{9-7}\)

\(B=\frac{6}{2}\)

\(B=3\)

vậy \(B=3\)

chiu ban oi

ta có: EM = \(\sqrt{OE^2-OM^2}=\sqrt{4R^2-R^2}=\sqrt{3}R.\)

(pytago)

xét tg vuông EOM và EDB có: góc E chung => 2 tam giác đồng dạng

=> \(\frac{EM}{OE}=\frac{EB}{ED}\Leftrightarrow ED=\frac{EB.OE}{EM}=\frac{2R.3R}{\sqrt{3}R}=2\sqrt{3}R\)

lại có: BD=\(\sqrt{ED^2-BE^2}=\sqrt{12R^2-9R^2}=\sqrt{3}R\)

(pytago)

xét tg vuông EAC và EBD có E chung => 2 tg đồng dạng

=>\(\frac{EA}{EB}=\frac{AC}{BD}\Leftrightarrow AC=\frac{EA.BD}{EB}=\frac{R.\sqrt{3}R}{3R}=\frac{R}{\sqrt{3}}\)

\(S_{ABDC}=\frac{\left(AC+DB\right).AB}{2}=\frac{\left(\frac{R}{\sqrt{3}}+\sqrt{3}R\right).2R}{2}=\left(3+\sqrt{3}\right).R^2\)

cảm ơn