Tính:
\(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\)
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\(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.24}+...+\frac{11}{89.100}\)
\(=\left(1-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{23}\right)+\left(\frac{1}{23}-\frac{1}{24}\right)+...+\left(\frac{1}{89}-\frac{1}{100}\right)\)
\(=1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
đúng thì thôi. sai thì khỏi
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(\frac{1}{1}-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{23}\right)+\left(\frac{1}{23}-\frac{1}{34}\right)+...+\left(\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(1-\frac{1}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\frac{99}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{3}-\frac{99}{100}\)
\(\Leftrightarrow\)\(x=\frac{203}{300}\)
Vậy \(x=\frac{203}{300}\)
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(1-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\frac{99}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{3}-\frac{99}{100}=\frac{203}{300}\)
\(\left(\frac{11}{1.12}+\frac{11}{12.23}+\frac{11}{23.34}+....+\frac{11}{89.100}\right).x=\frac{1}{100}\)
\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+.....+\frac{1}{98}-\frac{1}{100}\right)x=\frac{1}{100}\)
\(\left(1-\frac{1}{100}\right).x=\frac{1}{100}\)
\(\frac{99}{100}x=\frac{1}{100}\)
\(x=\frac{1}{100}:\frac{99}{100}\)
\(x=\frac{1}{99}\)
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{13.24}+...+\frac{11}{89.100}\right)+x=\frac{2}{3}\)
\(\Rightarrow\frac{11}{12}+\frac{11}{12}-\frac{11}{23}+...+\frac{11}{89}-\frac{11}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{1}{12}-\frac{1}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{11}{150}+x=\frac{2}{3}\)
=>\(x=\frac{2}{3}-\frac{11}{150}\)
=>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
\(\frac{11}{1.12}+\frac{11}{12.13}+\frac{11}{23.34}+...+\frac{11}{89.100}\)
\(=1-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-...-\frac{1}{89}+\frac{1}{89}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)