a)   3-4x\(\ge\)11

  \(4x\le3-11=-8\)

\(x\le-2\)

( câu b bn ghi rõ đề bài đc ko ?)

a: \(\Leftrightarrow\dfrac{3}{x-2}=\dfrac{2x-1}{x-2}-\dfrac{x\left(x-2\right)}{x-2}\)

=>3=2x-1-x^2+2x

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

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

=>x^2-4x+4=0

=>x=2(loại)

b: =>(x+2)(2x-4)=x(2x+3)

=>2x^2-4x+4x-8=2x^2+3x

=>3x=-8

=>x=-8/3(nhận)

a: 2x-3>3(x-2)

=>2x-3>3x-6

=>-x>-3

hay x<3

b: \(\dfrac{12x+1}{12}< =\dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)

=>12x+1<=36x+4-24x-3

=>12x+1<=12x+1(luôn đúng)

\(a,\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\\ \Rightarrow7.\left(2x-1\right)-3.\left(5x+2\right)=21.\left(x+13\right)\\ \Rightarrow14x-7-15x-6=21x+273\\\Rightarrow -x-21x=273+13\\ \Rightarrow-22x=286\\ \Rightarrow x=-13\\ b,\dfrac{3\left(x+3\right)}{4}+\dfrac{1}{2}=\dfrac{5x+9}{3}-\dfrac{7x-9}{4}=0\\ \Rightarrow9.\left(x+3\right)+6=4.\left(5x+9\right)-3.\left(7x-9\right)=0\\\Rightarrow 9x+27+6=20x+36-21x+27\\ \Rightarrow9x+33=-x+63\\ \Rightarrow10x=30\\ \Rightarrow x=3\)

\(a,\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)

\(\Rightarrow7\left(2x-1\right)-3\left(5x+2\right)-21x-273=0\)

\(\Rightarrow14x-7-15x-6-21x-273=0\)

\(\Rightarrow-22x=286\)

\(\Rightarrow x=-13\)

\(b,\dfrac{3\left(x+3\right)}{4}+\dfrac{1}{2}=\dfrac{5x+9}{3}-\dfrac{7x-9}{4}\)

\(\Rightarrow9\left(x+3\right)+6-4\left(5x+9\right)+3\left(7x-9\right)=0\)

\(\Rightarrow9x+27+6-20x-36+21x-27=0\)

\(\Rightarrow10x=30\Rightarrow x=3\)

a)

\(\dfrac{x-2}{4}+\dfrac{2x-3}{3}=\dfrac{x-18}{6}\)

`<=> 3x-6+8x-12=2x-36`

`<=> 3x+8x-2x=-36+6+12`

`<=> 9x=-18`

`<=> x=-2`

b)

\(\dfrac{x+3}{x-3}+\dfrac{3-x}{x+3}=\dfrac{36}{x^2-9}\left(x\ne3;x\ne-3\right)\)

suy ra

`(x+3)^2 +(3-x)(x-3)=36`

`<=>x^2 +6x+9+3x-9-x^2 +3x=36`

`<=> x^2 -x^2 +6x+3x+3x+9-9-36=0`

`<=> 12x-36=0`

`<=> 12x=36`

`<=> x=3 (KTMĐK)

a: =>x-2+2=x^2+2x

=>x^2+2x=x

=>x^2+x=0

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

b: =>-9(5x-8)+4(7x-12)=-6(x+18)

=>-45x+72+28x-48=-6x-108

=>-17x+24=-6x-108

=>-11x=-132

=>x=12

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

a: \(\Leftrightarrow\dfrac{y+5}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)

\(\Leftrightarrow2\left(y+5\right)^2-\left(y-5\right)^2=y^2+25y\)

=>\(2y^2+20y+50-y^2+10y-25=y^2+25y\)

=>30y+25=25y

=>5y=-25

=>y=-5(loại)

b: \(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)

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

=>2x^2-6x=0

=>2x(x-3)=0

=>x=0(nhận) hoặc x=3(loại)

c: =>x^2-9-6(2x+7)=-13(x+3)

=>x^2-9-12x-42+13x+39=0

=>x^2+x-6=0

=>(x+3)(x-2)=0

=>x=2(nhận) hoặc x=-3(loại)

1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4