Ta có: \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}\)\(=1\frac{98}{100}\)
\(\Leftrightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{1}{x.\left(x+1\right):2}\)
Lại có: \(\frac{1}{x\left(x+1\right):2}=\frac{1}{x\left(x+1\right)}.2\)\(=\frac{2}{x\left(x+1\right)}\)
\(\Rightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=1\frac{98}{100}\)
\(\Leftrightarrow2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=1\frac{98}{100}\)
\(\Leftrightarrow2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)\)\(=1\frac{98}{100}\)
\(\Leftrightarrow2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=1\frac{98}{100}\)
\(\Leftrightarrow2.\left(1-\frac{1}{x+1}\right)=1\frac{98}{100}\)
\(\Leftrightarrow1-\frac{1}{x+1}=1\frac{98}{100}:2\)\(=\frac{99}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
\(\Rightarrow x+1=100\)
\(\Leftrightarrow x=100-1=99\)
Vậy: x=99
Box toán dạo này chán quá ~~
