bay bài đó với

=>2+2/3+2/2x3+2/3x4+......+2/x*(x+1)=1989/1991

=>2+2/3+1/2-1/3+1/3-1/4+.......+1/x-1/x+1=1989/1991

=>tự tính nốt

sao nó dễ zữ vậy

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

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

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

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

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

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

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

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

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

Bạn nhân cả 2 vế với 1/2 là về dạng toán sai phân bình thường

\(2\left(1+\frac{1}{3}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{3980}{1991}\)

\(1+\frac{1}{3}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+......+\frac{x+1-x}{x\left(x+1\right)}=\frac{1990}{1991}\)

\(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}-\frac{1}{x-1}=\frac{1990}{1991}\)

\(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{x-1}=\frac{1990}{1991}\)

\(\frac{1}{x-1}=\frac{11}{6}-\frac{1990}{1991}=\frac{9961}{11946}\)

\(x-1=\frac{11946}{9961}\Rightarrow x=\frac{21907}{9961}\)