$t^2-t-(x^2+x)=0$ $\Delta =[2(x+\frac{1}{2})]^2$$t_1=1+2(x+\frac{1}{2})$$t_2=1-2(x+\frac{1}{2})$ Với $t=1+2(x+\frac{1}{2})=2x+2$PT <=>$3x^2+5x+2$ $x_1=-\frac{2}{3}$ $x_2=-1$TH2 tuong tu
$t^2-t-(x^2
-x)=0$ $\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