Question

Giải phương trình

\[
\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \ldots + \frac{1}{2023} - \left( \frac{1}{1013} + \frac{1}{1014} + \frac{1}{1015} + \ldots + \frac{1}{2024} \right) \cdot x = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots + \frac{1}{2024}
\]

Answer 1

\(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{2023}-\left(\dfrac{1}{1013}+\dfrac{1}{1014}+...+\dfrac{1}{2024}\right)\cdot x=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2024}\)

=>\(x\left(\dfrac{1}{1013}+\dfrac{1}{1014}+...+\dfrac{1}{2024}\right)=1+\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2}-\dfrac{1}{4}-...-\dfrac{1}{2024}\)

=>\(x\left(\dfrac{1}{1013}+\dfrac{1}{1014}+...+\dfrac{1}{2024}\right)=1+\dfrac{1}{2}+...+\dfrac{1}{2024}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2024}\right)\)

=>\(x\left(\dfrac{1}{1013}+\dfrac{1}{1014}+...+\dfrac{1}{2024}\right)=\dfrac{1}{1013}+\dfrac{1}{1014}+...+\dfrac{1}{2024}\)

=>x=1