Bạn kia làm đúng nên mk chỉ giải tiếp thôi!
\(\left(x-1\right)\left(x+3\right)\left(x^2-x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\\x^2-x+1=0\end{matrix}\right.\\ Vìx^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\Rightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\ Vậy...\)
\(x^4+x^3-4x^2+5x-3=0\)
\(\Leftrightarrow\left(x^4-x^3\right)+\left(2x^3-2x^2\right)-\left(2x^2-2x\right)+\left(3x-3\right)=0\)
\(\Leftrightarrow x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-2x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x^3+3x^2\right)-\left(x^2+3x\right)+\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+3\right)-x\left(x+3\right)+\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x^2-x+1\right)=0\)
Bí
