Bài 3:
a:Ta có: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=-\dfrac{1}{2}\\x+\dfrac{3}{4}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
b: Ta có: \(23-\left(2x+3\right)^2=-2\)
\(\Leftrightarrow\left(2x+3\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
Bài 2:
a: Ta có: \(\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
b: Ta có: \(x^3+4x=0\)
\(\Leftrightarrow x\left(x^2+4\right)=0\)
hay x=0