\(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.24}+...+\frac{11}{89.100}\)

\(=\left(1-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{23}\right)+\left(\frac{1}{23}-\frac{1}{24}\right)+...+\left(\frac{1}{89}-\frac{1}{100}\right)\)

\(=1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

đúng thì thôi. sai thì khỏi

TUI BÍT KB ĐI ZÙI LÀM

AND K NỮA

\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)

\(\Leftrightarrow\)\(\left(\frac{1}{1}-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{23}\right)+\left(\frac{1}{23}-\frac{1}{34}\right)+...+\left(\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)

\(\Leftrightarrow\)\(1-\frac{1}{100}+x=\frac{5}{3}\)

\(\Leftrightarrow\)\(\frac{99}{100}+x=\frac{5}{3}\)

\(\Leftrightarrow\)\(x=\frac{5}{3}-\frac{99}{100}\)

\(\Leftrightarrow\)\(x=\frac{203}{300}\)

Vậy \(x=\frac{203}{300}\)

\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)

\(\Leftrightarrow\)\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)

\(\Leftrightarrow\)\(\left(1-\frac{1}{100}\right)+x=\frac{5}{3}\)

\(\Leftrightarrow\)\(\frac{99}{100}+x=\frac{5}{3}\)

\(\Leftrightarrow\)\(x=\frac{5}{3}-\frac{99}{100}=\frac{203}{300}\)

1/11.(11/12+11/12-11/23+...+1/89-1/100)+x=5/3

1/11.(11/12+11/12-1/100)+x=5/3

1/11.547/300+x=5/3

547/3300+x=5/3

x=5/3-547/3300

x=1651/1100

\(\left(\frac{11}{1.12}+\frac{11}{12.23}+\frac{11}{23.34}+....+\frac{11}{89.100}\right).x=\frac{1}{100}\)

\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+.....+\frac{1}{98}-\frac{1}{100}\right)x=\frac{1}{100}\)

\(\left(1-\frac{1}{100}\right).x=\frac{1}{100}\)

\(\frac{99}{100}x=\frac{1}{100}\)

\(x=\frac{1}{100}:\frac{99}{100}\)

\(x=\frac{1}{99}\)

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

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

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

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

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

=>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 \(x=\dfrac{-97}{300}\)

có aj giải đc bài này ko

có đo tui nè

n oi sai de zui