Tìm x biết:
\(\left(\frac{11}{12}+\frac{11}{12.23}+...+\frac{11}{89.100}\right)+x=\frac{2}{3}\)
Hình như câu này sai đề
tìm x, biết
a, \(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.24}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
b, \(\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}\right)-x+4+\frac{221}{231}=\frac{7}{3}\)
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{13.24}+...+\frac{11}{89.100}\right)+x=\frac{2}{3}\)
Tính:
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)\)
\(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.24}+...+\frac{11}{89.100}\)
\(=\left(1-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{23}\right)+\left(\frac{1}{23}-\frac{1}{24}\right)+...+\left(\frac{1}{89}-\frac{1}{100}\right)\)
\(=1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
đúng thì thôi. sai thì khỏi
(\(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.24}+...+\frac{11}{89.100}\))+x = \(\frac{2}{3}\)
\(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.24}+...+\frac{11}{89.100}\)=?
Tính:
\(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\)
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
(\(\frac{11}{12}\)+\(\frac{11}{12.23}\)+\(\frac{11}{23.34}\)+....+\(\frac{11}{89.100}\)) + x = \(\frac{5}{3}\)
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(\frac{1}{1}-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{23}\right)+\left(\frac{1}{23}-\frac{1}{34}\right)+...+\left(\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(1-\frac{1}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\frac{99}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{3}-\frac{99}{100}\)
\(\Leftrightarrow\)\(x=\frac{203}{300}\)
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
Tìm X <- N
a) ( \(\frac{11}{12}\)+ \(\frac{11}{12.23}\)+ \(\frac{11}{23.34}\)+ ........ + \(\frac{11}{89.100}\)) . x = \(\frac{1}{100}\)
\(\left(\frac{11}{1.12}+\frac{11}{12.23}+\frac{11}{23.34}+....+\frac{11}{89.100}\right).x=\frac{1}{100}\)
\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+.....+\frac{1}{98}-\frac{1}{100}\right)x=\frac{1}{100}\)
\(\left(1-\frac{1}{100}\right).x=\frac{1}{100}\)
\(\frac{99}{100}x=\frac{1}{100}\)
\(x=\frac{1}{100}:\frac{99}{100}\)
\(x=\frac{1}{99}\)
Bài 2 : Tìm x, biết :
\(a.\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{15}{28}-\frac{11}{13}\right)\)
\(b.\left(\frac{7}{12}+x\right)-\frac{11}{12}=\frac{2}{3}\)