PT <=> \(\left(x^4-x^3\right)+\left(2x^3-2x^2\right)-\left(2x^2-2x\right)+\left(3x-3\right)=0\)
<=> \(x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)+3\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x^3+2x^2-2x+3\right)=0\)
<=> \(\left(x-1\right)\left[\left(x^3+3x^2\right)-\left(x^2+3x\right)+\left(x+3\right)\right]=0\)
<=> \(\left(x-1\right)\left[x^2\left(x+3\right)-x\left(x+3\right)+\left(x+3\right)\right]=0\)
<=> \(\left(x-1\right)\left(x+3\right)\left(x^2-x+1\right)=0\)
<=> \(\left[{}\begin{matrix}x-1=0\\x+3=0\\x^2-x+1=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=-3\\x=\varnothing\end{matrix}\right.\)
Lời giải:
PT $\Leftrightarrow x^4-x^3+2x^3-2x^2-2x^2+2x+3x-3=0$
$\Leftrightarrow x^3(x-1)+2x^2(x-1)-2x(x-1)+3(x-1)=0$
$\Leftrightarrow (x-1)(x^3+2x^2-2x+3)=0$
$\Leftrightarrow (x-1)(x^3+3x^2-x^2-3x+x+3)=0$
$\Leftrightarrow (x-1)[x^2(x+3)-x(x+3)+(x+3)]=0$
$\Leftrightarrow (x-1)(x+3)(x^2-x+1)=0$
Dễ thấy $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}>0$ nên $(x-1)(x+3)=0$
$\Rightarrow x=1$ hoặc $x=-3$
Vậy......