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

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-\dfrac{9}{2}\\x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{\left(3+\sqrt{9+2x}\right)^2.2x^2}{\left(3-\sqrt{9+2x}\right)^2\left(3+\sqrt{9+2x}\right)^2}< x+21\)

\(\Leftrightarrow\dfrac{\left(3+\sqrt{9+2x}\right)^2.2x^2}{4x^2}< x+21\)

\(\Leftrightarrow\left(3+\sqrt{9+2x}\right)^2< 2x+42\)

\(\Leftrightarrow x+9+3\sqrt{9+2x}< x+21\)

\(\Leftrightarrow\sqrt{9+2x}< 4\)

\(\Leftrightarrow9+2x< 16\Rightarrow x< \dfrac{7}{2}\)

Vậy \(\left\{{}\begin{matrix}-\dfrac{9}{2}\le x< \dfrac{7}{2}\\x\ne0\end{matrix}\right.\)

\(\dfrac{2x-1}{3}\)-\(\dfrac{x+3}{2}\)\(\le\)1

<=>\(\dfrac{2\left(2x-1\right)}{6}\)+\(\dfrac{3\left(x+3\right)}{6}\)\(\le\)\(\dfrac{6}{6}\)

=>4x -2 +3x+9\(\le\)6

<=>7x+7\(\le\)6

<=>7x\(\le\)6-7

<=>7x\(\le\)-1

<=>x\(\le\)\(\dfrac{-1}{7}\)

vậy bất phương trình có nghiệm là x\(\le\)\(\dfrac{-1}{7}\)

Mashiro ShiinaAkai HarumaNhã Doanh giúp mk vs

\(\dfrac{2x-1}{3}\)-\(\dfrac{x+3}{2}\)\(\le\)1

<=>\(\dfrac{2\left(2x-1\right)}{6}\)-\(\dfrac{3\left(x+3\right)}{6}\le\dfrac{6}{6}\)

=>4x-2-3x-9\(\le\)6

<=>x-11\(\le\)6

<=>x\(\le\)6+11

<=>x\(\le\)17

Vậy bất phương trình có nghiệm là x\(\le\)17