Tìm x biết biểu thức sau: 1/3 + 1/6 +1/10 + ..... + 2/x(x+1) = 2023/2025

\(\frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + \cdot \cdot \cdot + \frac{2}{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}} = \frac{{2023}}{{2025}}\)

\(\frac{2}{6} + \frac{2}{{12}} + \frac{2}{{20}} + \cdot \cdot \cdot + \frac{2}{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}} = \frac{{2023}}{{2025}}\)

\(2\left[ {\frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \cdot \cdot \cdot + \frac{1}{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}}} \right] = \frac{{2023}}{{2025}}\)

\(\frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + \cdot \cdot \cdot + \frac{1}{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}} = \frac{{2023}}{{4050}}\)

\(\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \cdot \cdot \cdot + \frac{1}{{\rm{x}}} - \frac{1}{{{\rm{x}} + 1}} = \frac{{2023}}{{4050}}\)

\(\frac{1}{2} - \frac{1}{{{\rm{x}} + 1}} = \frac{{2023}}{{4050}}\)

\(\frac{1}{{{\rm{x}} + 1}} = \frac{1}{2} - \frac{{2023}}{{4050}}\)

\(\frac{1}{{{\rm{x}} + 1}} = \frac{1}{{2025}}\)

x + 1 = 2025

x = 2024.

Vậy x = 2024.