a) 6 -9x = 0
=> 9x = 6
=> x = $\frac{6}{9}$= $\frac{2}{3}$
b)
$\frac{2x-3}{4}$+2= $\frac{1-x}{6}$
<=> $\frac{2x +5}{4}$= $\frac{1-x}{6}$
<=> $\frac{6x+15}{12}$ =$\frac{2-2x}{12}$
<=> $\frac{6x+15-2+2x}{12}$ =0
<=> 8x + 13 = 0
<=> 8x = -13
<=> x = $\frac{-13}{8}$
c) Ta có $x^2 +3$$\geq$ 3 >0 nên
8x -4 =0
=> 8x = 4
=> x= $\frac{1}{2}$