a: 2x-1>=5

nên 2x>=6

hay x>=3

b: \(\dfrac{x-2}{3}>=x-\dfrac{x-1}{2}\)

=>2x-4>=6x-3(x-1)

=>2x-4>=6x-3x+3

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

=>-x>=7

hay x<=-7

a.\(2x-1\ge5\)

\(\Leftrightarrow2x\ge6\)

\(\Leftrightarrow x\ge3\)

Vậy \(S=\left\{x|x\ge3\right\}\)

b.\(\dfrac{x-2}{3}\ge x-\dfrac{x-1}{2}\)

\(\Leftrightarrow\dfrac{2\left(x-2\right)}{6}\ge\dfrac{6x-3\left(x-1\right)}{6}\)

\(\Leftrightarrow2\left(x-2\right)\ge6x-3\left(x-1\right)\)

\(\Leftrightarrow2x-4\ge6x-3x+3\)

\(\Leftrightarrow-x\ge7\)

\(\Leftrightarrow x\le7\)

Vậy \(S=\left\{x|x\le7\right\}\)

d: =>-2x<=1

=>x>=-1/2

e: =>3x<=10

=>x<=10/3

f: =>2x-6+12<=x+2

=>2x+6<=x+2

=>x<=-4

`(x-2)/6 -(x-1)/3 < x/2`

`<=> (x-2)/6 -(2(x-1))/6 < (3x)/6`

`<=> x-2 - (2x-2) <3x`

`<=> x-2-2x+2<3x`

`<=> -x <3x`

`<=> -x-3x<0`

`<=> -4x<0`

`<=> x>0`

\(\dfrac{x-2}{6}\)-\(\dfrac{x-1}{3}\)<\(\dfrac{x}{2}\)

\(\dfrac{x-2}{6}\)-\(\dfrac{2\left(x-1\right)}{6}\)<\(\dfrac{6x}{6}\)

<=>x-2-2x+2<6x

<=>-7x<0

<=>x>0

vậy tập nghiệm của bất phương trình là

\(\left\{x|x>0\right\}\)

\(\dfrac{x-2}{6}-\dfrac{x-1}{3}< \dfrac{x}{2}\\ \Leftrightarrow\dfrac{x-2}{6}-\dfrac{\left(x-1\right)2}{3.2}< \dfrac{x.3}{2.3}\\ \Leftrightarrow x-2-2x+2< 3x\\ \Leftrightarrow x-2x-3x< -2\\ \Leftrightarrow-4x< -2\\ \Leftrightarrow x>\dfrac{1}{2}\)

Vậy bất phương trình có tập nghiệm là:

\(S=\left\{x|x>\dfrac{1}{2}\right\}\)

Biểu diễn:

a)

\(\dfrac{x-2}{2}+1\le\dfrac{x-1}{3}\)

\(\Leftrightarrow\dfrac{3\left(x-2\right)}{6}+\dfrac{1.6}{6}\le\dfrac{2\left(x-1\right)}{6}\)

`<=> 3x - 6 + 6 <= 2x-2`

`<=> 3x <= 2x-2`

`<=> 3x -2x <= -2`

`<=> x  <= -2`

\(\dfrac{x-2}{2}\)+1≤\(\dfrac{x-1}{3}\)

<=>\(\dfrac{3x-6}{6}\)+\(\dfrac{6}{6}\)≤\(\dfrac{2x-1}{6}\)

<=>3x-6+6≤2x-1

<=>x<-1

Ta có: \(x^2< 9\)

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

BPT \(\Leftrightarrow x>3\)

x+3>6 <=> x>3

-------0-----(3----->

\(x+3>6\)

\(\Leftrightarrow x>6-3\)

\(\Leftrightarrow x>3\)

Biểu diễn trên trục số:

\(x^2-4x+3\ge0\)

\(\left(x-1\right)\left(x-3\right)\ge0\)

TH1; X-1>=0 VA X-3>=0

TH2: X-1=<0 VA X-3<=0

Vay x>=3 hoac x<=1