Bài 1: 

a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)

\(\Leftrightarrow6-8x-10+2x-5=0\)

\(\Leftrightarrow-6x+11=0\)

\(\Leftrightarrow-6x=-11\)

hay \(x=\dfrac{11}{6}\)

b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)

\(\Leftrightarrow6-12x-11+3x-1=0\)

\(\Leftrightarrow-9x-6=0\)

\(\Leftrightarrow-9x=6\)

hay \(x=-\dfrac{2}{3}\)

   404 ERROR

AH SHIT

damnnnn this sus

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

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

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

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

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

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

Vậy \(S=\left\{3\right\}\)

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

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

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

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

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)

ĐKXĐ : \(x\ne\pm3\)

\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow-7x+3=-4x-15\)

\(\Leftrightarrow-7x+4x=-15-3\)

\(\Leftrightarrow-3x=-18\)

\(\Leftrightarrow x=6\)( tmđk )

Vậy x = 6 là nghiệm của phương trình

2) 2x + 3 < 6 - ( 3 - 4x )

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

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

<=> -2x < 0

<=> x > 0

Vậy nghiệm của bất phương trình là x > 0

Trước hết xoá \(\frac{2x}{a^2-a+1}\)ở 2 vế. Nếu \(\frac{a}{a+1}>0\left(a< -1;a>0\right)\)thì \(x< \frac{a}{4}\). Nếu \(\frac{a}{a+1}< 0\left(-1< a< 0\right)\)thì \(x>\frac{a}{4}\)

\(ĐKXĐ:a\ne-1\)

\(\frac{2x}{a^2-a+1}-\frac{1}{2a+2}< \frac{4x-1}{2a^2-2a+2}+\frac{a-2ax}{1+a^3}\Leftrightarrow\frac{2x}{a^2-a+1}-\frac{1}{2a+2}< \frac{2x}{a^2-a+1}-\frac{1}{2a^2-2a+2}+\frac{a}{1+a^3}-\frac{2ax}{1+a^3}\)\(\Leftrightarrow\frac{1}{2a+2}-\frac{1}{2a^2-2a+2}+\frac{a}{1+a^3}>\frac{2ax}{1+a^3}\Leftrightarrow\frac{a^2-a+1-a-1+2a}{2\left(a^3+1\right)}>\frac{2ax}{1+a^3}\Leftrightarrow\frac{a^2}{2\left(1+a^3\right)}>\frac{4ax}{2\left(1+a^3\right)}\)\(\Leftrightarrow\frac{4ax}{a+1}< \frac{a^2}{a+1}\)

* Nếu \(\frac{a}{a+1}>0\)(tức là a < -1 hoặc a > 0) thì \(x< \frac{a}{4}\)

* Nếu \(\frac{a}{a+1}< 0\)(tức là -1 < a < 0) thì \(x>\frac{a}{4}\)

?????????????????????????

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

d: =>3x>=9

=>x>=3