Có: \(\left(\left|x\right|-\dfrac{1}{8}\right).\left(-\dfrac{1}{5}\right)^5=\left(-\dfrac{1}{8}\right)^7\)
<=> \(\left|x\right|.\left(-\dfrac{1}{5}\right)^5-\dfrac{1}{8}.\left(-\dfrac{1}{5}\right)^5=\left(-\dfrac{1}{8}\right)^7\)
<=> \(\left|x\right|.\left(-\dfrac{1}{5}\right)^5+\dfrac{1}{8}.\left(\dfrac{1}{5}\right)^5=\left(-\dfrac{1}{8}\right)^7\)
<=> \(\left|x\right|.\dfrac{-1}{3125}=-\dfrac{1}{8^7}-\dfrac{1}{8}.\dfrac{1}{3125}\)
<=> \(\left|x\right|=\dfrac{\dfrac{-1.3125}{8^7.3125}-\dfrac{1}{8.3125}}{-\dfrac{1}{3125}}=\dfrac{\dfrac{-3125}{8^7}.\dfrac{1}{3125}-\dfrac{1}{8}.\dfrac{1}{3125}}{-\dfrac{1}{3125}}=\dfrac{\dfrac{-1}{3125}\left(\dfrac{3125}{8^7}+\dfrac{1}{8}\right)}{-\dfrac{1}{3125}}\)
<=> \(\left|x\right|=\dfrac{3125}{8^7}+\dfrac{8^6}{8^7}=\dfrac{265269}{2097152}\)
=> x\(\in\left\{\dfrac{265269}{2097152};\dfrac{-265269}{2097152}\right\}\)
