Bài 3:
a) 3(x-2)-4(x+3)=x-2
⇔ 3x-6-4x-12=x-2
⇔ 3x-6-4x-12-x+2=0
⇔-2x-16=0
⇔ -2(x+8)=0
⇔ $\left \{ {{-2=0} \atop {x+8=0}} \right.$
⇔ x=-8
b) 4x-5x²+7x-5+6x²=x-5
⇔ 4x-5x²+7x-5+6x²-x+5=0
⇔ 10x+x²=0
⇔ x(10+x)=0
⇔ $\left \{ {{x=0} \atop {10+x=0}} \right.$
⇔ $\left \{ {{x=0} \atop {x=-10}} \right.$