Mình nghĩ câu A phải là 11/23.34 mới đúng.

Theo đề mới: A)

11/12 = 11/1.12

Vậy A = 11/1.12 + 11/12.23 + ... + 11/89.100

= 1 - 1/12 + 1/12 - 1/23 + .... + 1/89 - 1/100

= 99/1000

Vậy x là: 2/3 - 99/100 = -97/300

B) 2/11.13 + ... + 2/19.21 = 1/11 - 1/13 + .... + 1/19 - 1/21 = 10/231

=> 10/231 - x = 4/3 - 221/223 = 229/669

=> x = 10/231 - 229/669

=> x = 6690/154539 - 52899/154539

=> x = -46209/154539 = -15403/51513

Giúp mik luôn bài chứng minh đi bạn T.T

a.

\(\left(\frac{11}{12}+\frac{11}{12\times23}+\frac{11}{23\times34}+...+\frac{11}{89\times100}\right)+x=\frac{2}{3}\)

\(\left(\frac{11}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{2}{3}\)

\(\left(\frac{11}{12}+\frac{1}{12}-\frac{1}{100}\right)+x=\frac{2}{3}\)

\(\left(\frac{12}{12}-\frac{1}{100}\right)+x=\frac{2}{3}\)

\(\left(1-\frac{1}{100}\right)+x=\frac{2}{3}\)

\(\left(\frac{100-1}{100}\right)+x=\frac{2}{3}\)

\(\frac{99}{100}+x=\frac{2}{3}\)

\(x=\frac{2}{3}-\frac{99}{100}\)

\(x=\frac{200-297}{300}\)

\(x=-\frac{97}{300}\)

b.

\(\left(\frac{2}{11\times13}+\frac{2}{13\times15}+...+\frac{2}{19\times21}\right)-x+\frac{221}{231}=\frac{4}{3}\)

\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)+\frac{221}{231}-x=\frac{4}{3}\)

\(\left(\frac{1}{11}-\frac{1}{21}\right)+\frac{221}{231}-x=\frac{4}{3}\)

\(\left(\frac{21-11}{231}\right)+\frac{221}{231}-x=\frac{4}{3}\)

\(\frac{10}{231}+\frac{221}{231}-x=\frac{4}{3}\)

\(\frac{231}{231}-x=\frac{4}{3}\)

\(1-x=\frac{4}{3}\)

\(x=1-\frac{4}{3}\)

\(x=\frac{3-4}{3}\)

\(x=-\frac{1}{3}\)

Chúc bạn học tốt

có aj giải đc bài này ko

có đo tui nè

n oi sai de zui

\(\Rightarrow\frac{11}{12}+\frac{11}{12}-\frac{11}{23}+...+\frac{11}{89}-\frac{11}{100}+x=\frac{2}{3}\)

\(\Rightarrow\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}+x=\frac{2}{3}\)

\(\Rightarrow\frac{1}{12}-\frac{1}{100}+x=\frac{2}{3}\)

\(\Rightarrow\frac{11}{150}+x=\frac{2}{3}\)

=>\(x=\frac{2}{3}-\frac{11}{150}\)

=>x=\(\frac{89}{150}\)

Ta có:

\(\left(\dfrac{11}{12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{2}{3}\)

\(\Leftrightarrow\left(\dfrac{11}{1.12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{2}{3}\)

\(\Leftrightarrow\left(1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+...+\dfrac{1}{89}-\dfrac{1}{100}\right)+x=\dfrac{2}{3}\)

\(\Leftrightarrow\left(1-\dfrac{1}{100}\right)+x=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{99}{100}+x=\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{2}{3}-\dfrac{99}{100}=\dfrac{-97}{300}\)

Vậy \(x=\dfrac{-97}{300}\)