a) 3x-2+x=23-x  

<=> 3x+x+x=23+2

<=> 5x=25

<=>x  = 5

Vậy pt có no duy nhất x=5

 b) 4-(3x-6)=10

<=>4-3x+6=10

<=>-3x=10-6-4

<=>-3x=0

 <=>x=0

 Vậy...

c) 3x$\frac{x}{2}$ + $\frac{2x-1}{3}$ = 2

<=> $\frac{3x}{6}$ + $\frac{2(2x-1)}{6}$ = $\frac{12}{6}$

<=> 3x+2(2x-1)=12

<=> 3x+4x-2=12

<=> 3x+4x= 12+2

<=> 7x=14

<=> x=2

Vậy...

d)$\frac{1-2x}{4}$ -$\frac{1-5x}{8}$ =1

<=>$\frac{2(1-2x)}{8}$-$\frac{1-5x}{8}$= $\frac{8}{8}$ 

<=> 2(1-2x)-1-5x=8

<=> 2-4x-1-5x=8

<=> -4x-5x=8-2+1

<=>-9x=7

<=> x=$\frac{-7}{9}$

Vậy...

e)  (2x+1)(1-3x)=0

<=> 2x-$6x^{2}$ +1-3x=0

<=>2x-$6x^{2}$ -3x=-1 

<=>$6^{2}$ -x=1( vô lý)

Vậy pt vô nghiệm

f) $\frac{x-1}{x+2}$ -$\frac{x+1}{x-2}$ =$\frac{x-3}{4-x^{2}}$ 

<=>$\frac{(x-2)(x-1)}{(x-2)(x+2)}$ -$\frac{x+2)(x+1)}{(x-2)(x+2)}$ = $\frac{-x+3}{(x-2)(x+2)}$ 

<=> (x-2)(x-1)-(x+2)(x+1)=-x+3

<=> $x^{2}$ -x-2x+2-$x^{2}$ -x-2x-2=-x+3

<=>$x^{2}$-x-2x $x^{2}$ -x-2x+x=3-2+2

<=>-5x=3

<=>x=$\frac{-3}{5}$ 

Vậy...

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