$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})Vớit=1+2(x+\frac{1}{2})=2x+2PT<=>3x^2+5x+2x_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})
Vớit=1+2(x
-\frac{1}{2})=2x
PT<=>3x^2
-x
=0x_1=\frac{
1}{3}
x_2=
0$TH2 tuong tu