Question

[Hoàng Bảo Trân]

Giải phương trình :\(\left(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+......+\frac{3}{97\cdot99}\right)\left(2x+1\right)=x+\frac{1}{33}\)

Answer 1

[Stephen Hawking]

\(\left(\frac{3}{1.3}+\frac{3}{3.5}+.......+\frac{3}{97.99}\right).\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow[\frac{3}{2}.(\frac{2}{1.3}+\frac{2}{3.5}+.......+\frac{2}{97.99})].\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow[\frac{3}{2}.(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{97}-\frac{1}{99})].\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow[\frac{3}{2}.(1-\frac{1}{99})].\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow\left(\frac{3}{2}.\frac{98}{99}\right).\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow\frac{49}{33}.\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow\frac{49}{33}.2x+\frac{49}{33}=x+\frac{1}{33}\)

\(\Rightarrow\frac{98}{33}.x+\frac{49}{33}=x+\frac{1}{33}\)

\(\Rightarrow\frac{98}{33}.x-x=\frac{1}{33}-\frac{49}{33}\)

\(\Rightarrow\frac{65}{33}.x=\frac{-16}{11}\)

\(\Rightarrow x=\frac{-16}{11}:\frac{65}{33}\)

\(\Rightarrow x=\frac{-48}{65}\)

Vậy \(x=\frac{-48}{65}\)