Giải bất phương trình sau và biểu diễn tập nghiệm trên trục số:
\(\dfrac{1-2x}{4}-2< \dfrac{1-5x}{8}\)
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a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
\(\Leftrightarrow3\left(1-2x\right)-2\left(x+1\right)< =6\)
=>3-6x-2x-2<=6
=>-8x+1<=6
=>-8x<=5
hay x>=5/8
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
\(\dfrac{x-2}{2}+1\le\dfrac{x-1}{3}\)
\(\Leftrightarrow\dfrac{3\left(x-2\right)}{6}+\dfrac{1.6}{6}\le\dfrac{2\left(x-1\right)}{6}\)
`<=> 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
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<
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 \(x\le-6\)
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
\(\dfrac{1-2x}{4}-2< \dfrac{1-5x}{8}\)
\(\Leftrightarrow\dfrac{2\left(1-2x\right)-16}{8}< \dfrac{1-5x}{8}\)
\(\Leftrightarrow2\left(1-2x\right)-16< 1-5x\)
\(\Leftrightarrow2-4x-16< 1-5x\)
\(\Leftrightarrow x< 15\)
Vậy \(S=\left\{x|x< 15\right\}\)
nhanh qá=')