Question

1) Giải BPT

a) \(\left(4x+1\right)^2\ge1\)

b) \(|2x+3|\ge\dfrac{1}{2}\)

Answer 1

a/

\(\left(4x+1\right)^2\ge1\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+1\ge1\\4x+1\le-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\le-\dfrac{1}{2}\end{matrix}\right.\)

Vậy.....

b/ \(\left|2x+3\right|\ge\dfrac{1}{2}\)\(\Leftrightarrow\left[{}\begin{matrix}2x+3\ge\dfrac{1}{2}\\2x+3\le-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge-\dfrac{5}{4}\\x\le-\dfrac{7}{4}\end{matrix}\right.\)

Vậy......