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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x-2}{2}+1\le\dfrac{x-1}{3}\)
\(\Leftrightarrow\dfrac{3\left(x-2\right)}{6}+\dfrac{1.6}{6}\le\dfrac{2\left(x-1\right)}{6}\)
`<=> 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
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(S=\left\{x|x\le7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x\le-6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{x+2}{x-3}< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2>0\\x-3< 0\end{matrix}\right.\Leftrightarrow-2< x< 3\)
Vậy: S={x|-2<x<3}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow3\left(1-2x\right)-2\left(x+1\right)< =6\)
=>3-6x-2x-2<=6
=>-8x+1<=6
=>-8x<=5
hay x>=5/8
![](https://rs.olm.vn/images/avt/0.png?1311)
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
`(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`
Cảm ơn bạn nhìu nha