$\begin{cases}|x^2-5x+4|>x-1\\x>1\\\end{cases}$

$\to \begin{cases}(x^2-5x+4)^2>(x-1)^2\\x>1\\\end{cases}$

$\to \begin{cases}(x-1)^2(x-4)^2>(x-1)^2\\x>1\\\end{cases}$

$\to \begin{cases}(x-1)^2[(x-4)^2-1]>0\\x>1\\\end{cases}$

$\to \begin{cases}(x-4)^2-1>0\\x>1\\\end{cases}$

$\to \begin{cases}(x-5)(x-3)>0\\x>1\\\end{cases}$

$\to \begin{cases}\left[ \begin{array}{l}x>5\\x<3\end{array} \right.\\x>1\\\end{cases}$

$\to \left[ \begin{array}{l}1<x<3\\x>5\end{array} \right.$

Vậy bất phương trình có tập nghiệm $S=(1,3]∩(5,∞]$

Giải từng bất phương trình bằng cách chuyển vế rồi lập bảng xét dấu là ra nha bạn

a) 3(x-2)(x+2) < 3x2 + x

3(x² + 2x - 2x - 4 ) < 3x² + x

<=> 3x² + 6x - 6x - 12 < 3x² + x

<=> 3x² + 6x - 6x - 3x² - x < 12

<=> x > -12

Vậy bpt có nghiệm là x > -12.

b) ( x+4 )(5x-1) > 5x² + 16x + 2

<=> 5x² - x + 20x - 4 - 5x² - 16x - 2 > 0

<=> 5x² - x + 20x - 5x² - 16x > 2 + 4

<=> 3x > 6

<=> x > 2

Vậy btp có nghiệm là x > 2

Giải:

a) \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)

\(\Leftrightarrow3\left(x^2-4\right)< 3x^2+x\)

\(\Leftrightarrow3x^2-12< 3x^2+x\)

\(\Leftrightarrow-12< x\)

\(\Leftrightarrow x>-12\)

Vậy ...

b) \(\left(x+4\right)\left(5x-1\right)>5x^2+16x+2\)

\(\Leftrightarrow5x^2+20x-x-4>5x^2+16x+2\)

\(\Leftrightarrow5x^2+19x-4>5x^2+16x+2\)

\(\Leftrightarrow3x-4>2\)

\(\Leftrightarrow3x>6\)

\(\Leftrightarrow x>2\)

Vậy ...

a: \(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)< 10x\left(2x+3\right)-100\)

\(\Leftrightarrow20x^2-12x+15x-5< 20x^2+30x-100\)

=>3x-5<=30x-100

=>30x-100>3x-5

=>27x>95

hay x>95/27

b: \(\Leftrightarrow4\left(5x-2\right)-6\left(2x^2-x\right)< 4x\left(1-3x\right)-15x\)

\(\Leftrightarrow20x-8-12x^2+6x< 4x-12x^2-15x\)

=>26x-8<-11x

=>37x<8

hay x<8/37

a: \(\Leftrightarrow4\left(4x-2\right)+12\left(-x+3\right)< =3\left(1-5x\right)\)

=>16x-8-12x+36<=3-15x

=>4x+28<=3-15x

=>19x<=-25

hay x<=-25/19

b: \(\Leftrightarrow6\left(x+4\right)+30\left(-x-5\right)>=10\left(x+3\right)-15\left(x-2\right)\)

=>6x+24-30x-150<=10x+30-15x+30

=>-24x-126<=-5x+60

=>-19x<=186

hay x>=-186/19

\(a,\dfrac{4x-2}{3}-x+3\le\dfrac{1-5x}{4}\\ \Leftrightarrow\dfrac{4\left(4x-2\right)}{12}-\dfrac{12\left(x-3\right)}{12}\le\dfrac{3\left(1-5x\right)}{12}\\ \Leftrightarrow16x-8-12x+36\le3-15x\\ \Leftrightarrow4x+28\le3-15x\\ \Leftrightarrow19x+25\le0\\ \Leftrightarrow x\le-\dfrac{25}{19}\)

\(b,\dfrac{x+4}{5}-x-5\ge\dfrac{x+3}{3}-\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x+5\right)}{30}\ge\dfrac{10\left(x+3\right)}{30}-\dfrac{15\left(x-2\right)}{30}\\ \Leftrightarrow6x+24-30x-150\ge10x+30-15x+30\\ \Leftrightarrow-24x-126\ge-5x+60\\ \Leftrightarrow19x+186\le0\\ \Leftrightarrow x\le-\dfrac{186}{19}\)