Bạn tham khảo :
f )
$(\dfrac{2}{3}x + \dfrac{1}{3})^3 + \dfrac{27}{64} = 0$
⇒ $(\dfrac{2}{3}x + \dfrac{1}{3})^3 = \dfrac{-27}{64}$
⇒ $(\dfrac{2}{3}x + \dfrac{1}{3})^3 = (\dfrac{-3}{4})^3$
⇒ $(\dfrac{2}{3}x + \dfrac{1}{3})= \dfrac{-3}{4}$
⇒ $\dfrac{2}{3}x = \dfrac{-13}{12}$
⇒ $x = \dfrac{-13}{8}$
g )
$(x+ \dfrac{1}{4})^4 = \dfrac{1}{16}$
⇒ $(x+\dfrac{1}{4})^4 = (\dfrac{1}{2})^4$
⇒ $x+ \dfrac{1}{4} = \dfrac{±1}{2}$
⇒ $x ∈$ { $\dfrac{1}{4} ; \dfrac{-3}{4}$}
h )
$(3x - 15)^5 = -1024$
⇒ $(3x-15)^5 = (-4)^5$
⇒ $3x - 15= (-4)$
⇒ $x = \dfrac{3}{11}$
i )
$( \dfrac{5}{2}x - \dfrac{1}{2})^2 ) - \dfrac{49}{25} = 0$
⇒ $(\dfrac{5}{2}x - \dfrac{1}{2})^2 = \dfrac{49}{25}$
⇒$(\dfrac{5}{2}x - \dfrac{1}{2})^2 =± (\dfrac{7}{5})^2$
⇒ $\dfrac{5}{2}x - \dfrac{1}{2} = ±\dfrac{7}{5}$
⇒ $x ∈${$\dfrac{19}{25} ; \dfrac{-9}{25}$}
j )
$(\dfrac{1}{3}x + \dfrac{15}{27})^3 + \dfrac{36}{27} = \dfrac{100}{27}$
⇒ $(\dfrac{1}{3}x + \dfrac{15}{27})^3 = \dfrac{64}{27} = (\dfrac{4}{3})^3$
⇒ $\dfrac{1}{3}x + \dfrac{15}{27} = \dfrac{4}{3}$
⇒ $x = \dfrac{7}{3}$
