a. 5-(x-6)=4(3-2x)
⇔5-x+6=12-8x
⇔7x=1 ⇔ x=$\frac{1}{7}$
b. $\frac{4x-2}{3}$-x+1=$\frac{1-5x}{4}$
⇔$\frac{16x-8}{12}$- $\frac{12x}{12}$+12= $\frac{3-15x}{12}$
⇔16x-8-12x+12=3-15x ⇔ 9x=-1 ⇔ x=-1/9
c. x(x+1)-(x-2)(x+3)=3x+1
⇔x²+x-x²-x+6=3x+1 ⇔ 3x+1=6 ⇔ x=5/3
d. $\frac{x-1}{x+3}$ - $\frac{x}{x-3}$= $\frac{7x-3}{9-x^{2} }$ (1)
TXĐ: x≠±3
(1)⇔$\frac{x-1}{x+3}$+ $\frac{x}{3-x}$= $\frac{7x-3}{9-x^{2}}$
⇔(x-1)(3-x)+x(3+x)=7x-3
⇔ 4x-x²-3+x²+3x=7x-3 (VN)
e. l2x-1l =5 ⇔$\left \{ {{2x-1=5} \atop {2x-1=-5}} \right.$
⇔$\left \{ {{x=3} \atop {x=-2}} \right.$
