Sửa đề : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+....+\frac{\left(x+1\right)-x}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{1}-\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{50}\)
\(\Leftrightarrow x=50-1=49\)
Sửa đề: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
(=) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
(=)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{98}{100}\)
(=)\(1-\frac{1}{x+1}=\frac{98}{100}\)
(=)\(\frac{1}{x+1}=1-\frac{98}{100}\)
(=)\(\frac{1}{x+1}=\frac{1}{50}\)=> \(x+1=50\)
\(x=50-1\)
\(x=49\)
T_i_c_k cho mình nha,thanks you so much!
