x3+(x3(x−1)3+3x2x−1)−2=0⇔(x2−2x+2)(x4−x3+2x2−2x+1)(x−1)3=0⇔x2−2x+2=0hoặc:x4−x3+2x2−2x+1=0mà :x2−2x+2=(x−1)2+1>0(vôno)con:$x^4-x^3+2x^2-2x+1= 4(x^4-x^3+\frac{x^2}{4})+7x^2-8x+1=4(x^2-\frac{x}{2})^2+(x-\frac{4}{7})^2+\frac{12}{7}>0$tóm lại pt vô no
x3+(x3(x−1)3+3x2x−1)−2=0⇔(x2−2x+2)(x4−x3+2x2−2x+1)(x−1)3=0⇔x2−2x+2=0hoặc
:x4−x3+2x2−2x+1=0mà
:x2−2x+2=(x−1)2+1>0(vôno)con:$x^4-x^3+2x^2-2x+1= 4(x^4-x^3+\frac{x^2}{4})+7x^2-8x+
4=4(x^2-\frac{x}{2})^2+(x-\frac{4}{7})^2+\frac{12}{7}>0$tóm lại pt vô no