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

\(ĐK:x\ge\frac{2}{3}\)

PT \(\Leftrightarrow\sqrt{3x-2}-\frac{\sqrt{10}}{2}-\sqrt{x+1}+\frac{\sqrt{10}}{2}=\left(2x-3\right)\left(x+1\right)\)

\(\Leftrightarrow\frac{3x-2-\frac{5}{2}}{\sqrt{3x-2}+\frac{\sqrt{10}}{2}}-\frac{x+1-\frac{5}{2}}{\sqrt{x+1}+\frac{\sqrt{10}}{2}}-\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\frac{3x-\frac{9}{2}}{\sqrt{3x-2}+\frac{\sqrt{10}}{2}}+\frac{x-\frac{3}{2}}{\sqrt{x+1}+\frac{\sqrt{10}}{2}}-2\left(x-\frac{3}{2}\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)\left(\frac{3}{\sqrt{3x+2}+\frac{\sqrt{10}}{2}}+\frac{1}{x+1+\frac{\sqrt{10}}{2}}-2\left(x+1\right)\right)=0\)

\(\Rightarrow x=\frac{3}{2}\)(TMĐKXĐ)