g: Ta có: \(\left(x+5\right)^3=-27\)
\(\Leftrightarrow x+5=-3\)
hay x=-8
h: Ta có: \(\left(2x-3\right)^3=-64\)
\(\Leftrightarrow2x-3=-4\)
\(\Leftrightarrow2x=-1\)
hay \(x=-\dfrac{1}{2}\)
a) \(\Rightarrow\dfrac{2}{5}x=-\dfrac{3}{20}\Rightarrow x=-\dfrac{3}{8}\)
b) \(\Rightarrow\dfrac{1}{2x}=-\dfrac{3}{5}\Rightarrow x=-\dfrac{5}{6}\)
c) \(\Rightarrow\dfrac{-2}{3}:x=-\dfrac{29}{24}\Rightarrow x=\dfrac{16}{29}\)
d) \(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{3}{4}\\x+\dfrac{1}{2}=-\dfrac{3}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
e) \(\Rightarrow\dfrac{1}{6}x=\dfrac{1}{15}\Rightarrow x=\dfrac{2}{5}\)
f) \(\Rightarrow\dfrac{11}{15}x=-\dfrac{2}{5}\Rightarrow x=-\dfrac{6}{11}\)
g) \(\Rightarrow x+5=-3\Rightarrow x=-8\)
h) \(\Rightarrow2x-3=-4\Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
k) \(\Rightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
