A, 3X+6>0

(=)3X>-6

(=)X>-2

VẬY ...

B,10-2X≥-4

(=)-2X≥-4-10

(=)-2X≥-14

(=)X≤7

VẬY....

C,

(=)

(=) -15X+10>-3+3X

(=)-15X-3X>-3-10

(=)-18X>-13

(=)X<

Ở câu (h) mình quên gạch chân phân số, bạn thông cảm nha <3

1: =>2(x+2)>3x+1

=>2x+4-3x-1>0

=>-x+3>0

=>-x>-3

=>x<3

2: =>12x^2-2x>12x^2+9x-8x-6

=>-2x>-x-6

=>-x>-6

=>x<6

3: =>4(x+1)-12>=3(x-2)

=>4x+4-12>=3x-6

=>4x-8>=3x-6

=>x>=2

4: =>-5x<=15

=>x>=-3

5: =>3(x+2)-5(x-2)<30

=>3x+6-5x+10<30

=>-2x+16<30

=>-2x<14

=>x>-7

6: =>5(x+2)<3(3-2x)

=>5x+10<9-6x

=>11x<-1

=>x<-1/11

\(\dfrac{20x-20}{5}\)-\(\dfrac{2-x}{15}\)<\(\dfrac{30x-9}{15}\)

=20x-20-2+x<30x-9

=21x-22<30x-9

điều này đúng vs mọi x

10) \(\dfrac{8x+12}{28}\)>\(\dfrac{7x-35}{28}\)

    =>8x+12>x-5 (đpcm)

điều này đúng vs mọi x

g: =>12x+1>=36x+12-24x-3

=>12x+1>=12x+9(loại)

h: =>6(x-1)+4(2-x)<=3(3x-3)

=>6x-6+8-4x<=9x-9

=>2x+2<=9x-9

=>-7x<=-11

=>x>=11/7

i: =>4x^2-12x+9>4x^2-3x

=>-12x+9>-3x

=>-9x>-9

=>x<1

1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)

=>-4x<12

hay x>-3

2: \(\Leftrightarrow6+2x+2>2x-1-12\)

=>8>-13(đúng)

4: \(\dfrac{2x+1}{x-3}\le2\)

\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)

=>x-3<0

hay x<3

6: =>(x+4)(x-1)<=0

=>-4<=x<=1

2:

a: =>x-4>=0

=>x>=4

b: =>x+1>0

=>x>-1

d: Ta có: \(\dfrac{2x+1}{3}-\dfrac{1-x}{2}\ge1-\dfrac{x}{4}\)

\(\Leftrightarrow8x+4-6+6x\ge12-3x\)

\(\Leftrightarrow14x+3x\ge12+2=14\)

\(\Leftrightarrow x\ge\dfrac{14}{17}\)

e: Ta có: \(\dfrac{x+1}{2}-\dfrac{2-x}{3}< \dfrac{2x-3}{4}\)

\(\Leftrightarrow6x+12+4x-8< 6x-9\)

\(\Leftrightarrow4x< -9+8-12=-13\)

hay \(x< -\dfrac{13}{4}\)

\(\dfrac{2x-3}{2}>\dfrac{8x-11}{6}\)

\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{6}>\dfrac{8x-11}{6}\)

\(\Leftrightarrow3\left(2x-3\right)>8x-11\)

\(\Leftrightarrow6x-9>8x-11\)

\(\Leftrightarrow-2x>-2\)

\(\Leftrightarrow x< 1\)

Vậy \(S=\left\{x|x< 1\right\}\)

\(2x-3\le8x-11\)

\(\Leftrightarrow-6x\le-8\)

\(\Leftrightarrow x\ge\dfrac{8}{6}\)

Vậy \(S=\left\{x|x\ge\dfrac{8}{6}\right\}\)

\(\dfrac{x-3}{2}>\dfrac{x-11}{3}\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{6}>\dfrac{2\left(x-11\right)}{6}\)

\(\Leftrightarrow3\left(x-3\right)>2\left(x-11\right)\)

\(\Leftrightarrow3x-9>2x-22\)

\(\Leftrightarrow x>-13\)

Vậy \(S=\left\{x|x>-13\right\}\)

Ta có: \(\dfrac{x-1}{3}-\dfrac{3x+5}{2}\ge1-\dfrac{4x+5}{6}\)

\(\Leftrightarrow2\left(x-1\right)-3\left(3x+5\right)\ge6-4x-5\)

\(\Leftrightarrow2x-2-9x-15-6+4x+5\ge0\)

\(\Leftrightarrow-3x\ge18\)

hay \(x\le-6\)