11/12.23+11/23.34=...+11/89.100
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TN
3 tháng 4 2016
\(\Rightarrow\frac{11}{12}+\frac{11}{12}-\frac{11}{23}+...+\frac{11}{89}-\frac{11}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{1}{12}-\frac{1}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{11}{150}+x=\frac{2}{3}\)
=>\(x=\frac{2}{3}-\frac{11}{150}\)
=>x=\(\frac{89}{150}\)
21 tháng 3 2017
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 \(x=\dfrac{-97}{300}\)
MS
0
NT
2
DT
3
22 tháng 3 2018
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 ~
LK
22 tháng 3 2018
chắc bạn đánh thiếu đề
\(\frac{11}{1.12}+\frac{11}{12.13}+\frac{11}{23.34}+...+\frac{11}{89.100}\)
\(=1-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-...-\frac{1}{89}+\frac{1}{89}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
6 tháng 3 2017
\(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.24}+...+\frac{11}{89.100}\)
\(=\left(1-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{23}\right)+\left(\frac{1}{23}-\frac{1}{24}\right)+...+\left(\frac{1}{89}-\frac{1}{100}\right)\)
\(=1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
đúng thì thôi. sai thì khỏi
27 tháng 6 2018
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) \(\frac{x+1}{9}+\frac{x+3}{7}+\frac{x+5}{5}+\frac{x+7}{3}=a\)
\(\Rightarrow\frac{x+1}{9}+1+\frac{x+3}{7}+1+\frac{x+5}{5}+1+\frac{x+7}{3}+1=a+4\)
\(\frac{x+10}{9}+\frac{x+10}{7}+\frac{x+10}{5}+\frac{x+10}{3}=a+4\)
\(\left(x+10\right).\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=a+4\)
27 tháng 6 2018
\(a,\left(\frac{11}{12}+\frac{11}{12\cdot23}+\frac{11}{23\cdot34}+...+\frac{11}{89\cdot100}\right)-x=\frac{2}{3}\)
\(\Rightarrow\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\Rightarrow1-\frac{1}{100}-x=\frac{2}{3}\)
\(\Rightarrow\frac{99}{100}-x=\frac{2}{3}\)
\(\Rightarrow x=\frac{99}{100}-\frac{2}{3}\)
\(\Rightarrow x=\frac{97}{300}\)
b, k hiểu đề :v
23 tháng 6 2022
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
30 tháng 1 2018
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(\frac{1}{1}-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{23}\right)+\left(\frac{1}{23}-\frac{1}{34}\right)+...+\left(\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(1-\frac{1}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\frac{99}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{3}-\frac{99}{100}\)
\(\Leftrightarrow\)\(x=\frac{203}{300}\)
Vậy \(x=\frac{203}{300}\)
KT
30 tháng 1 2018
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(1-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\frac{99}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{3}-\frac{99}{100}=\frac{203}{300}\)
\(\frac{11}{12.23}+\frac{11}{23.24}+....+\frac{11}{89.100}\)
\(=\frac{11}{12}-\frac{11}{23}+\frac{11}{23}-\frac{11}{24}+....+\frac{11}{89}-\frac{11}{100}\)
\(=\frac{11}{12}-\frac{11}{100}\)
\(=\frac{68}{75}\)
=1/12-1/23+1/23-1/34+...+1/89-1/100
=1/12-1/100
=121/150