$x^3+(\frac{x^3}{(x-1)^3}+\frac{3x^2}{x-1})-2=0$$\Leftrightarrow \frac{(x^2-2x+2)(x^4-x^3+2x^2-2x+1)}{(x-1)^3}=0$$\Leftrightarrow x^2-2x+2=0$hoặc$:x^4-x^3+2x^2-2x+1=0$mà $:x^2-2x+2=(x-1)^2+1>0(vô no)$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
$x^3+(\frac{x^3}{(x-1)^3}+\frac{3x^2}{x-1})-2=0$$\Leftrightarrow \frac{(x^2-2x+2)(x^4-x^3+2x^2-2x+1)}{(x-1)^3}=0$$\Leftrightarrow x^2-2x+2=0$hoặc$:x^4-x^3+2x^2-2x+1=0$mà $:x^2-2x+2=(x-1)^2+1>0(vô no)$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
$x^3+(\frac{x^3}{(x-1)^3}+\frac{3x^2}{x-1})-2=0$$\Leftrightarrow \frac{(x^2-2x+2)(x^4-x^3+2x^2-2x+1)}{(x-1)^3}=0$$\Leftrightarrow x^2-2x+2=0$hoặc$:x^4-x^3+2x^2-2x+1=0$mà $:x^2-2x+2=(x-1)^2+1>0(vô no)$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