Question

Giúp mình 2 bài này với ạ

Answer 1

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: \(23-\left(2x+3\right)^2=-2\)

\(\Leftrightarrow\left(2x+3\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

Answer 2

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