10.
为解方程\({(}x^{2}{-}1{{)}}^{2}{-}5(x^{2}{-}1){+}4{=}0\),我们可以将\(x^{2}{-}1\)视为一个整体,然后设\(x^{2}{-}1{=}y{,}\)则\({(}x^{2}{-}1{{)}}^{2}{=}y^{2}{,}\)原方程化为\(y^{2}{-}5y{+}4{=}0{,}\)解得\(y_{1}{=}1{,}y_{2}{=}4{.}\)
当\(y=1\)时,\(x^{2}{-}1{=}1{,}\)得\(x{=}{±}\sqrt{2}\);当\(y=4\)时,\(x^{2}{-}1{=}4{,}\)得\(x{=}{±}\sqrt{5}.\) 故原方程的解为\(x_{1}{=}\sqrt{2}{,}x_{2}{=}{-}\sqrt{2}{,}x_{3}{=}\sqrt{5}{,}x_{4}{=}{-}\sqrt{5}{.}\)在解方程的过程中,我们将\(x^{2}{-}1\)用\(y\)替换,先解出关于\(y\)的方程,达到了降低方程次数的目的,这种方法叫做“换元法”,体现了转化的数学思想。
请根据以上的阅读,解下列方程:
\((1)x^{4}{-}x^{2}{-}6{=}0\)
\((2){(}\dfrac{1}{2}x{-}1{{)}}^{2}{-}(\dfrac{1}{2}x{-}1){-}1{=}0\)