Đáp án:
Giải thích các bước giải:
a) x(1-2x)+(2x²-3x+4) = 0
⇔x-2x²+2x²-3x+4=0
⇔-2x+4=0
⇔x=2
b)x.(x-5) -(x-5).(3x+2)=0
⇔(x-5).[(x-(3x+2)] = 0
⇔(x-5).(-2x-2)=0
⇔$\left \{ {{x-5=0} \atop {-2x-2=0}} \right.$
⇔$\left \{ {{x=5} \atop {x=-1}} \right.$
c) 2($\frac{1}{3}$x-$\frac{1}{2}$ x)-$\frac{1}{2}$ (3-x)=0
⇔$\frac{2}{3}$x-1-$\frac{3}{2}$+$\frac{1}{2}$x=0
⇔$\frac{7}{6}$x-$\frac{5}{2}$=0
⇔x=$\frac{15}{4}$
d) ($\frac{1}{3}$x-1)-($\frac{4}{3}$x-2)+(-$\frac{1}{2}$-4)-8=0
⇔$\frac{1}{3}$x-$\frac{4}{3}$x-1+2-$\frac{1}{2}$-4-8=0
⇔-x+($\frac{-23}{2}$=0
⇔x=$\frac{-23}{2}$
