\(\left(x^2+5\right)\left(2x+3\right)\left(3x-1\right)< 0\)

Do \(\left(x^2+5\right)>0\)

\(\Rightarrow bpt\Leftrightarrow\left(2x+3\right)\left(3x-1\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+3>0\\3x-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}2x+3< 0\\3x-1>0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\frac{-3}{2}\\x< \frac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \frac{-3}{2}\\x>\frac{1}{3}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\frac{-3}{2}< x< \frac{1}{3}\left(chon\right)\\\frac{1}{3}< x< \frac{-3}{2}\left(loai\right)\end{matrix}\right.\)

Vậy...

(2x-1)(x-3)-3x+1≤(x-1)(x+3)+x²-5

<=> 2x²-6x-x+3-3x+1≤x²+3x-x-3+x²-5

<=> -12x≤-6

<=>x≥\(\frac{1}{2}\)

Vậy nghiệm của bpt là S=[\(\frac{1}{2}\);+∞)

3x-5>-2x+5

⇔ 3x+2x > 5+5

⇔ 5x >5

⇔ x>1

vậy bpt có tập nghiệm là S={ x/ x>1}

\(\Leftrightarrow\left|2x-1\right|< 3x+5\Leftrightarrow\left[{}\begin{matrix}2x-1< 3x+5\\1-2x< 3x+5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-3x< 5+1\\-2x-3x< 5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x< 6\\-5x< 4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-6\\x>-\frac{4}{5}\end{matrix}\right.\)

Bài 1: 

c) |2x - 1| = x + 2

<=> 2x - 1 = +(x + 2) hoặc -(x + 2)

* 2x - 1 = x + 2      

<=> 2x - x = 2 + 1

<=> x = 3

* 2x - 1 = -(x + 2)

<=> 2x - 1 = x - 2

<=> 2x - x = -2 + 1

<=> x = -1

Vậy.....

2x-x(3x+1)≤15-3x(x+2)

2x-3x²-x≤15-3x² -6x

2x-3x²-x+3x² +6x≤15

7x≤15

x≤15/7

ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\x\ge\frac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x-1>\sqrt{2x^2-3x-5}\)

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

- Với \(x\ge\frac{5}{2}\) hai vế ko âm, bình phương:

\(x^2-2x+1>2x^2-3x-5\)

\(\Leftrightarrow x^2-x-6< 0\Rightarrow-2< x< 3\)

\(\Rightarrow\frac{5}{2}\le x< 3\)