a,(2x+3)2=\(\frac{9}{21}\)
2x+3=
b) \(\left(3x-1\right)^3=-\frac{8}{27}\)
\(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-1=-\frac{2}{3}\)
\(3x=-\frac{2}{3}+1\)
\(3x=\frac{1}{3}\)
\(x=\frac{1}{3}:3\)
\(x=\frac{1}{3}.\frac{1}{3}\)
\(x=\frac{1}{9}\)
Vậy x = 1/9
c) |2x + 1| - 3 = 6
|2x + 1| = 6 + 3
|2x + 1| = 9
=> |2x + 1| ∈ {9;-9}
\(\Rightarrow\hept{\begin{cases}2x+1=9\\2x+1=-9\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x=9-1\\2x=-9-1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x=8\\2x=-10\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=8:2=4\\x=-10:2=-5\end{cases}}\)
Vậy x ∈ {4;-5}