ĐKXĐ: \(-2\le x\le3\)

\(\Leftrightarrow3x^3+3x^2-12x-12+x+4-3\sqrt{x+2}+5-x-3\sqrt{3-x}\ge0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left(3x+6\right)+\frac{x^2-x-2}{x+4+3\sqrt{x+2}}+\frac{x^2-x-2}{5-x+3\sqrt{3-x}}\ge0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left[3\left(x+2\right)+\frac{1}{x+4+3\sqrt{x+2}}+\frac{1}{5-x+3\sqrt{3-x}}\right]\ge0\)

\(\Leftrightarrow x^2-x-2\ge0\)

\(\Rightarrow\left[{}\begin{matrix}-2\le x\le-1\\2\le x\le3\end{matrix}\right.\)

ĐKXĐ: \(1\le x\le3\)

- Với \(1\le x< 2\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT luôn đúng

- Với \(x\ge2\) hai vế ko âm, bình phương:

\(-x^2+4x-3>x^2-4x+4\)

\(\Leftrightarrow2x^2-8x+7< 0\Rightarrow2\le x< \frac{4+\sqrt{2}}{2}\)

Vậy nghiệm của BPT là: \(1\le x< \frac{4+\sqrt{2}}{2}\)

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

x-3 ; mình đánh thiếu

ĐKXĐ: \(x\ge\frac{1}{4}\)

\(\sqrt{5x+1}\le3\sqrt{x}+\sqrt{4x-1}\)

\(\Leftrightarrow5x+1\le9x+4x-1+6\sqrt{4x^2-x}\)

\(\Leftrightarrow3\sqrt{4x^2-x}\ge1-4x\)

Do \(x\ge1\Rightarrow\left\{{}\begin{matrix}1-4x\le0\\\sqrt{4x^2-x}\ge0\end{matrix}\right.\) \(\Rightarrow\) BPT luôn đúng

Vậy nghiệm của BPT là \(x\ge\frac{1}{4}\)

b/ ĐKXĐ: \(x\ge4\)

\(\Leftrightarrow\sqrt{2\left(x^2-16\right)}+x-3>7-x\)

\(\Leftrightarrow\sqrt{2\left(x^2-16\right)}>10-2x\)

- Với \(x>5\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT luôn đúng

- Với \(x\le5\) bình phương 2 vế:

\(2\left(x^2-16\right)>4\left(x-5\right)^2\)

\(\Leftrightarrow x^2-20x+66< 0\)

\(\Rightarrow10-\sqrt{34}< x< 10+\sqrt{34}\)

Vậy nghiệm của BPT là \(x>10-\sqrt{34}\)

\(x^2-4x-21>0\)

\(\Leftrightarrow\)  \(x^2-4x+4>25\)

\(\Leftrightarrow\) \(\left(x-2\right)^2>25\)

\(\Leftrightarrow\) \(\left|x-2\right|>5\)

\(\Leftrightarrow\orbr{\begin{cases}x-2>5\\x-2>-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x>7\\x>-3\end{cases}}}\)

\(x^2-4x-21>0\)

\(x^2-4x+4-25>0\)

\(\left(x-2\right)^2>25\)

Ta có: \(25=5^2=\left(-5\right)^2\)

TH1: \(\left(x-2\right)^2>5^2\)

\(x-2>5\)

\(x>7\)

TH2: \(\left(x-2\right)^2>\left(-5\right)^2\)

\(x-2>-5\)

\(x>-3\)

Kết hợp cả 2 TH ta đc x>-3

=.= hok tốt!!