a, \(\left|3x+1\right|>2\)

\(\Leftrightarrow\left(\left|3x+1\right|\right)^2>4\)

\(\Leftrightarrow9x^2+6x+1>4\)

\(\Leftrightarrow9x^2+6x-3>0\)

\(\Leftrightarrow3\left(3x-1\right)\left(x+1\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< -1\end{matrix}\right.\)

b, \(\left|2x-1\right|\le1\)

\(\Leftrightarrow\left(\left|2x-1\right|\right)^2\le1\)

\(\Leftrightarrow4x^2-4x+1\le1\)

\(\Leftrightarrow4x\left(x-1\right)\le0\)

\(\Leftrightarrow0\le x\le1\)

c, ĐK: \(x\ne13\)

\(\left|\dfrac{2}{x-13}\right|>\dfrac{8}{9}\)

\(\Leftrightarrow\dfrac{1}{\left|x-13\right|}>\dfrac{4}{9}\)

\(\Leftrightarrow4\left|x-13\right|< 9\)

\(\Leftrightarrow16\left(x^2-26x+169\right)< 81\)

\(\Leftrightarrow16x^2-416x+2623< 0\)

\(\Leftrightarrow\dfrac{43}{4}< x< \dfrac{61}{4}\)

\(\Rightarrow\) Có hai giả trị thỏa mãn yêu cầu bài toán

Bài 1 :

Ta có : \(\dfrac{3x+5}{2}-1\le\dfrac{x+2}{3}+x\)

\(\Leftrightarrow\dfrac{3x+5}{2}-1-\dfrac{x+2}{3}-x\le0\)

\(\Leftrightarrow\dfrac{3\left(3x+5\right)-6-2\left(x+2\right)-6x}{6}\le0\)

\(\Leftrightarrow9x+15-6-2x-4-6x\le0\)

\(\Leftrightarrow x\le-5\)

Mà \(\left\{{}\begin{matrix}x\in Z\\x>-10\end{matrix}\right.\)

Vậy \(x\in\left\{-5;-6;-7;-8;-9\right\}\)

b3\(\Leftrightarrow2x^2+5x-3-3x+1\le x^2+2x-3+x^2-5\\ \Leftrightarrow0.x\le-6\Leftrightarrow x\in\varnothing\)

a, \(\left|4x-8\right|\le8\)

\(\Leftrightarrow\left(\left|4x-8\right|\right)^2\le64\)

\(\Leftrightarrow16x^2-64x+64\le64\)

\(\Leftrightarrow16x^2-64x\le0\)

\(\Leftrightarrow16x\left(x-4\right)\le0\)

\(\Leftrightarrow0\le x\le4\)

b, \(\left|x-5\right|\le4\)

\(\Leftrightarrow\left(\left|x-5\right|\right)^2\le16\)

\(\Leftrightarrow x^2-10x+25\le16\)

\(\Leftrightarrow x^2-10x+9\le0\)

\(\Leftrightarrow1\le x\le9\)

\(\Rightarrow x\in\left\{1;2;3;4;5;6;7;8;9\right\}\)

c, \(\left|2x+1\right|< 3x\)

TH1: \(x\ge-\dfrac{1}{2}\)

\(\left|2x+1\right|< 3x\)

\(\Leftrightarrow2x+1< 3x\)

\(\Leftrightarrow x>1\)

\(\Rightarrow\left\{{}\begin{matrix}x\in Z\\x\in\left(1;2018\right)\end{matrix}\right.\)

TH2: \(x< -\dfrac{1}{2}\)

\(\left|2x+1\right|< 3x\)

\(\Leftrightarrow-2x-1< 3x\)

\(\Leftrightarrow x>-\dfrac{1}{5}\left(l\right)\)

Vậy \(\left\{{}\begin{matrix}x\in Z\\x\in\left(1;2018\right)\end{matrix}\right.\)

d, \(\left|x+1\right|+\left|x\right|< 3\)

\(\Leftrightarrow x+1+x+2\left|x^2+x\right|< 9\)

\(\Leftrightarrow\left|x^2+x\right|< 4-x\)

Xét hai trường hợp để phá dấu giá trị tuyệt đối

e, Tương tự câu d

a) 4x -8 ≥ 3(3x-1)-2x +1

⇒4x -8 ≥7x -2

⇒4x -7x ≥ -2 +8

⇒-3x ≥ 6

⇒x≤-2

Vậy bpt có nghiệm là:{x|x≤-2}

b) (x-3)(x+2)+(x+4)²≤ 2x (x+5)+4

⇔ x²+2x - 3x - 6 +x² + 8x +16≤ 2x² + 10x +4

⇔ x² +2x - 3x + x² + 8x - 2x²- 10x ≤ 4+6-16

⇔ -3x ≤ -6

⇔ x≥ 2

Vậy bpt có tập nghiệm là: {x|x≥2}

e: =>-3<5x-12<3

=>9<5x<15

=>9/5<x<3

f: =>3x+15>=3 hoặc 3x+15<=-3

=>3x>=-12 hoặc 3x<=-18

=>x<=-6 hoặc x>=-4

b: =>(2x-7)(x-5)<=0

=>7/2<=x<=5

1:

c: =>1/3x+2/3-x+1>x+3

=>-2/3x+5/3-x-3>0

=>-5/3x-4/3>0

=>-5x-4>0

=>x<-4/5

d: =>3/2x+5/2-1<=1/3x+2/3+x

=>3/2x+3/2<=4/3x+2/3

=>1/6x<=2/3-3/2=-5/6

=>x<=-5

2:

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9