Lời giải:
\(S=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2025})-(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2024})\\ =(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2025})-2(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2024})\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2025}-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1012}\\ =\frac{1}{1013}+\frac{1}{1014}+...+\frac{1}{2025}\\ =P\)
$\Rightarrow (S-P)^{2025}=0^{2025}=0$
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