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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
a/ \(\left(2x-3\right)\left(3x-4\right)\left(5x+2\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}-\frac{2}{3}< x< \frac{4}{3}\\x>\frac{3}{2}\end{matrix}\right.\)
b/ \(\Leftrightarrow24x^2-10x-25< 0\)
\(\Rightarrow-\frac{5}{6}< x< \frac{5}{4}\)
c/ \(\frac{4x\left(3x+2\right)}{2x+5}>0\Rightarrow\left[{}\begin{matrix}-\frac{5}{2}< x< -\frac{2}{3}\\x>0\end{matrix}\right.\)
d/ \(\Leftrightarrow\frac{3x+2}{2x-5}-\frac{2x-5}{3x+2}\ge0\)
\(\Leftrightarrow\frac{\left(3x+2\right)^2-\left(2x-5\right)^2}{\left(2x-5\right)\left(3x+2\right)}\ge0\)
\(\Leftrightarrow\frac{\left(5x-2\right)\left(x+7\right)}{\left(2x-5\right)\left(3x+2\right)}\ge0\Rightarrow\left[{}\begin{matrix}x\le-7\\-\frac{2}{3}< x\le\frac{2}{5}\\x>\frac{5}{2}\end{matrix}\right.\)
ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\Leftrightarrow4x^2\le\left(2x+9\right)\left(2x+2-2\sqrt{1+2x}\right)\)
\(\Leftrightarrow4x^2\le4x^2+22x+18-2\left(2x+9\right)\sqrt{2x+1}\)
\(\Leftrightarrow22x+18-2\left(2x+9\right)\sqrt{2x+1}\ge0\)
Đặt \(\sqrt{2x+1}=t\ge0\Rightarrow2x=t^2-1\)
\(11\left(t^2-1\right)+18-2\left(t^2+8\right)t\ge0\)
\(\Leftrightarrow2t^3-11t^2+16t-7\le0\)
\(\Leftrightarrow\left(t-1\right)^2\left(2t-7\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t\le\frac{7}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{2x+1}=1\\\sqrt{2x+1}\le\frac{7}{2}\end{matrix}\right.\)
\(\Rightarrow-\frac{1}{2}\le x\le\frac{45}{8}\)