ĐKXĐ: \(x\ge-\frac{3}{2}\)

Do \(1+\sqrt{3+2x}>0\) nên BPT tương đương:

\(4\left(x+1\right)^2\left(1+\sqrt{3+2x}\right)^2< \left(2x+1\right)\left(1-\sqrt{3+2x}\right)^2\left(1+\sqrt{3+2x}\right)^2\)

\(\Leftrightarrow4\left(x+1\right)^2\left(1+\sqrt{3+2x}\right)^2< \left(2x+1\right).4\left(x+1\right)^2\)

- Với \(x=-1\) ko phải là nghiệm

- Với \(x\ne-1\)

\(\Leftrightarrow\left(1+\sqrt{3+2x}\right)^2< 2x+1\)

\(\Leftrightarrow4+2x+2\sqrt{3+2x}< 2x+1\)

\(\Leftrightarrow2\sqrt{3+2x}< -3\)

BPT vô nghiệm

Gt ⇔ \(\left|2x-3\right|\le x+1\)

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3\le x+1\\x\ge\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}3-2x\le x+1\\x< \dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

 ⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le4\\x\ge\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\x< \dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\dfrac{3}{2}\le x\le4\\\dfrac{2}{3}\le x< \dfrac{3}{2}\end{matrix}\right.\)

⇔ \(\dfrac{2}{3}\le x\le4\)

Vậy bất phương trình có tập nghiệm là

\(S=\left[\dfrac{2}{3};4\right]\)