Question

Giúp với nhé !

Answer 1

Bài 1:

a)\(\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{111.113}\)

\(=\frac{1}{2}\cdot\left(\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{111.113}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{111}-\frac{1}{113}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{7}-\frac{1}{113}\right)\)

\(=\frac{1}{2}\cdot\frac{106}{791}=\frac{53}{791}\)

b)\(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{53\cdot55}\)

\(=\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\)

\(=\frac{1}{11}-\frac{1}{55}=\frac{4}{55}\)

Bài 2:

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)

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

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

\(\Rightarrow x+1=100\Rightarrow x=99\)

Answer 2

Bài 1 :

a) \(\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{111.113}\)

= \(\frac{1}{2}.\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{111.113}\right)\)

= \(\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{111}-\frac{1}{113}\right)\)

= \(\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{113}\right)\)

= \(\frac{1}{2}.\frac{106}{791}=\frac{53}{791}\)

b) \(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{53.55}\)

= \(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\)

= \(\frac{1}{11}-\frac{1}{55}=\frac{4}{55}\)

Answer 3

Bài 1:
a) \(\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{111.113}\)

\(=\frac{1}{2}\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{111.113}\right)\)

\(=\frac{1}{2}\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{111}-\frac{1}{113}\right)\)

\(=\frac{1}{2}\left(\frac{1}{7}-\frac{1}{113}\right)\)

\(=\frac{1}{2}.\frac{106}{791}\)

\(=\frac{53}{791}\)

b) \(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{53.55}\)

\(=\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\)

\(=\frac{1}{11}-\frac{1}{55}\)

\(=\frac{4}{55}\)

Bài 2:
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{100}\)

\(\Rightarrow x+1=100\)

\(\Rightarrow x=99\)

Vậy \(x=99\)