\(VP=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot5}+\dfrac{1}{5\cdot13}+\dfrac{1}{13\cdot8}+\dfrac{1}{8\cdot19}+\dfrac{1}{19\cdot11}+\dfrac{1}{11\cdot25}\\ =\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+\dfrac{2}{10\cdot13}+\dfrac{2}{13\cdot16}+\dfrac{2}{16\cdot19}+\dfrac{2}{19\cdot22}+\dfrac{2}{22\cdot25}\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}+\dfrac{3}{16\cdot19}+\dfrac{3}{19\cdot22}+\dfrac{3}{22\cdot25}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{25}\right)\\ =\dfrac{2}{3}\cdot\left(1-\dfrac{1}{25}\right)\\ =\dfrac{2}{3}\cdot\dfrac{24}{25}\\ =\dfrac{16}{25}\)\(a^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\\ a^2-\dfrac{9}{25}=\dfrac{16}{25}\\ a^2=\dfrac{16}{25}+\dfrac{9}{25}\\ a^2=1\\ \Rightarrow\left[{}\begin{matrix}a=1\\a=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-a=-1\\-a=1\end{matrix}\right.\)
Vậy \(-a=-1\) hoặc \(-a=1\)