Question

[Bùi Ngọc Trường Sơn]

Tìm x 

\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+......+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)

Answer 1

[Khánh Ngọc]

\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)

\(2\left(1+\frac{1}{3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\frac{1989}{1991}\)

\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)

\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)

\(\frac{8}{3}+2-\frac{2}{x+1}=1\frac{1989}{1991}\)

\(\frac{2}{x+1}=\frac{13}{10}\)( số thập phân dài quá nên mk lấy số tròn thôi nha )

\(x+1=2:\frac{13}{10}\)

\(x+1=\frac{20}{13}\)

\(\Leftrightarrow x=\frac{7}{13}\)