a)AB=OA-OB=7-4=3cm

b)I là trung điểm OB=>OI=BI=4/2=2cm

=> EI=OE+OI=2+2=4cm

Bạn muốn hỏi gì?

@Nguyễn Đức Huy, bạn ấn vào đọc tiếp thì xem đc nhé!

\(\dfrac{1}{4}\left(\dfrac{2}{5}x-\dfrac{1}{4}\right)-\dfrac{3}{4}=-\dfrac{3}{2}\)

=>\(\dfrac{1}{4}\left(\dfrac{2}{5}x-\dfrac{1}{4}\right)=-\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{-3}{4}\)

=>\(\dfrac{2}{5}x-\dfrac{1}{4}=-3\)

=>\(\dfrac{2}{5}x=-3+\dfrac{1}{4}=\dfrac{-11}{4}\)

=>\(x=-\dfrac{11}{4}:\dfrac{2}{5}=\dfrac{-11}{4}\cdot\dfrac{5}{2}=\dfrac{-55}{8}\)

\(\dfrac{1}{4}\cdot\left(\dfrac{2}{5}x-\dfrac{1}{4}\right)-\dfrac{3}{4}=-\dfrac{3}{2}\)

\(\Rightarrow\dfrac{1}{4}\cdot\left(\dfrac{2}{5}x-\dfrac{1}{4}\right)=-\dfrac{3}{2}+\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{4}\cdot\left(\dfrac{2}{5}x-\dfrac{1}{4}\right)=\dfrac{-3}{4}\)

\(\Rightarrow\dfrac{2}{5}x-\dfrac{1}{4}=\dfrac{-3}{4}\cdot4\)

\(\Rightarrow\dfrac{2}{5}x-\dfrac{1}{4}=-3\)

\(\Rightarrow\dfrac{2}{5}x=-3+\dfrac{1}{4}\)

\(\Rightarrow\dfrac{2}{5}x=\dfrac{-35}{4}\)

\(\Rightarrow x=\dfrac{-11}{4}:\dfrac{2}{5}\)

\(\Rightarrow x=\dfrac{-55}{8}\)

\(\dfrac{3}{5}+\dfrac{3}{20}+\dfrac{3}{44}+\dfrac{3}{77}\)

\(=\left(\dfrac{3}{5}+\dfrac{3}{20}\right)+\left(\dfrac{3}{44}+\dfrac{3}{77}\right)\)

\(=\left(\dfrac{12}{20}+\dfrac{3}{20}\right)+\left(\dfrac{21}{308}+\dfrac{12}{308}\right)\)

\(=\dfrac{15}{20}+\dfrac{33}{308}\)

\(=\dfrac{3}{4}+\dfrac{3}{28}\)

\(=\dfrac{21}{28}+\dfrac{3}{28}\)

\(=\dfrac{24}{28}\)

\(=\dfrac{6}{7}\)

Cảm ơn anh nhìu nheee🤞

Là sao ạ? Bạn có thể miêu tả kĩ hơn đc ko?

Số nghịch đảo của `-7/3` là `-3/7.`

\(\dfrac{7}{-3}\)

Đặt \(A=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)

\(\dfrac{1}{5^2}< \dfrac{1}{4\cdot5}=\dfrac{1}{4}-\dfrac{1}{5}\)

...

\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}=\dfrac{1}{99}-\dfrac{1}{100}\)

Do đó: \(A=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=>\(A< \dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\)