$t^2-t-(x^2-x)=0$ $\Delta =[2(x-\frac{1}{2})]^2$$t_1=1+2(x-\frac{1}{2})$$t_2=1-2(x-\frac{1}{2})$ Với $t=1+2(x-\frac{1}{2})=2x$PT <=>$3x^2-x=0$ $x_1=\frac{1}{3}$ $x_2=0$TH2 tuong tu
$t^2-t-(x^2-x)=0$ $\Delta =[2(x-\frac{1}{2})]^2$$t_1=\frac{1+2(x-\frac{1}{2})}{2}=x$$t_2=\frac{1-2(x-\frac{1}{2})}{2}=1-x$