\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\\ \Leftrightarrow3A=3\left(+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\right)\\ =1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

Lấy 3A - A ta được
\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\right)\\ 2A=1-\dfrac{1}{3^{100}}\\ \Leftrightarrow A=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)

\(E=\sqrt{37-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{3}-\sqrt{10}\right)^2}\\ =\left|3\sqrt{3}-\sqrt{10}\right|\\ =3\sqrt{3}-\sqrt{10}\)

\(F=\sqrt{51-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{5}-\sqrt{6}\right)^2}\\ =\left|3\sqrt{5}-\sqrt{6}\right|\\ =3\sqrt{5}-\sqrt{6}\)

\(G=\sqrt{59-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{6}-\sqrt{5}\right)^2}\\ =\left|3\sqrt{6}-\sqrt{5}\right|\\ =3\sqrt{6}-\sqrt{5}\)

\(H=\sqrt{17-2\sqrt{30}}\\ =\sqrt{\left(\sqrt{15}-\sqrt{2}\right)^2}\\ =\left|\sqrt{15}-\sqrt{2}\right|=\sqrt{15}-\sqrt{2}\)

Tứ giác ABCD có \(\widehat{A}+\widehat{D}=180^O\) 

=> AB // CD

=> tg ABCD là hình thang

lại có \(\widehat{C}=\widehat{D}=70^o\)

=> tứ giác ABCD là hình thang cân