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

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

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

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

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

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

Ta có : 

\(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\)

\(=\)\(\frac{11}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\)

\(=\)\(\frac{11}{12}+\frac{1}{12}-\frac{1}{100}\)

\(=\)\(\frac{12}{12}-\frac{1}{100}\)

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

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

Vậy \(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\)

Chúc bạn học tốt ~ 

chắc bạn đánh thiếu đề

\(\frac{11}{1.12}+\frac{11}{12.13}+\frac{11}{23.34}+...+\frac{11}{89.100}\)

\(=1-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-...-\frac{1}{89}+\frac{1}{89}-\frac{1}{100}\)

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

=11/12+11/12-11/23+11/23-11/24+...+11/89-11/100

=11/100

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

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

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

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

đúng thì thôi. sai thì khỏi

TUI BÍT KB ĐI ZÙI LÀM

AND K NỮA

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

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

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

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

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

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

b) \(\frac{x+1}{9}+\frac{x+3}{7}+\frac{x+5}{5}+\frac{x+7}{3}=a\)

\(\Rightarrow\frac{x+1}{9}+1+\frac{x+3}{7}+1+\frac{x+5}{5}+1+\frac{x+7}{3}+1=a+4\)

\(\frac{x+10}{9}+\frac{x+10}{7}+\frac{x+10}{5}+\frac{x+10}{3}=a+4\)

\(\left(x+10\right).\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=a+4\)

\(a,\left(\frac{11}{12}+\frac{11}{12\cdot23}+\frac{11}{23\cdot34}+...+\frac{11}{89\cdot100}\right)-x=\frac{2}{3}\)

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

\(\Rightarrow1-\frac{1}{100}-x=\frac{2}{3}\)

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

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

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

b, k hiểu đề :v

a) ( 11 12 + 11 12.23 + 11 23.34 + . . . + 11 89.100 ) − x = 2 3 ( 12 11 + 12.23 11 + 23.34 11 +...+ 89.100 11 )−x= 3 2 ( 1 − 1 12 + 1 12 − 1 23 + 1 23 − 1 34 + . . . + 1 89 − 1 100 ) − x = 2 3 (1− 12 1 + 12 1 − 23 1 + 23 1 − 34 1 +...+ 89 1 − 100 1 )−x= 3 2 ( 1 − 1 100 ) − x = 2 3 (1− 100 1 )−x= 3 2 99 100 − x = 2 3 100 99 −x= 3 2 x = 99 100 − 2 3 x= 100 99 − 3 2 x = 97 300 x= 300 97 b) x + 1 9 + x + 3 7 + x + 5 5 + x + 7 3 = a 9 x+1 + 7 x+3 + 5 x+5 + 3 x+7 =a ⇒ x + 1 9 + 1 + x + 3 7 + 1 + x + 5 5 + 1 + x + 7 3 + 1 = a + 4 ⇒ 9 x+1 +1+ 7 x+3 +1+ 5 x+5 +1+ 3 x+7 +1=a+4 x + 10 9 + x + 10 7 + x + 10 5 + x + 10 3 = a + 4 9 x+10 + 7 x+10 + 5 x+10 + 3 x+10 =a+4 ( x + 10 ) . ( 1 9 + 1 7 + 1 5 + 1 3 ) = a + 4 (x+10).( 9 1 + 7 1 + 5 1 + 3 1 )=a+4

b: \(\Leftrightarrow1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}+x=\dfrac{5}{3}\)

=>x+99/100=5/3

=>x=5/3-99/100=203/300

c: \(\Leftrightarrow\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)

\(\Leftrightarrow\dfrac{10}{231}-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)

=>5-x=7/3

hay x=8/3

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

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

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

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

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

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

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