\(x^3-x^2+x-2x^2+2x-2\) =0

\(x\left(x^2-x+1\right)-2\left(x^2-x+1\right)=0\)

\(\left(x-2\right)\left(x^2-x+1\right)\)=0

\(\left\{{}\begin{matrix}x=2\\x^2-x+1=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=2\\x=\dfrac{1\pm}{3}\end{matrix}\right.\sqrt{3}}\)

a. \(5x^2-6x+1\)

=\(5x^2-x-5x+1\)

\(=\left(x-1\right)\left(5x-1\right)\)

a. $5x^2-6x+1$

=$5x^2-x-5x+1$

$=\left(x-1\right)\left(5x-1\right)$

c

cBài 1:Thực hiện phép tính
\(a,2x^2y\left(x^2+xy-3y^2\right)\)

\(=2x^4y+2x^3y^2-6x^2y^3\)

\(b,x^2\left(2x^3-4x+3\right)\)

\(=2x^5-4x^3+3x^2\)

\(c,\left(3x+4x^2-2\right)\left(-x^2+1\right)\)

Ta có x + y + z = 0 

<=> (x + y + z)² = 0

<=> \(x^2+y^2+z^2+2xy+2yz+2zx=0\)

\(\Leftrightarrow xy+yz+zx=-3\) (vì x² + y² + z² = 6)

\(\Leftrightarrow x\left(y+z\right)+yz=-3\)

\(\Leftrightarrow-x^2+yz=-3\Leftrightarrow yz=x^2-3\) (vì x + y + z = 0)

Khi đó \(x^3+y^3+z^3=x^3+(y+z).(y^2+z^2-yz)\)

\(=x^3-x.[6-x^2-(x^2-3)]\)

\(=x^3-x.(9-2x^2)=3x^3-9x=6\)

Ta được \(\Leftrightarrow x^3-3x-2=0\Leftrightarrow(x^3+1)-3(x+1)=0\)

\(\Leftrightarrow(x+1)(x^2-x-2)=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Với x = -1 ta có hệ \(\left\{{}\begin{matrix}y+z=1\\y^2+z^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-z\\(1-z)^2+z^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-z\\z^2-z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-z\\\left[{}\begin{matrix}z=-1\\z=2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2\\z=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1\\z=2\end{matrix}\right.\end{matrix}\right.\)

Với x = 2 ta có hệ : \(\left\{{}\begin{matrix}y+z=-2\\y^2+z^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2-z\\(-2-z)^2+z^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2-z\\z^2+2z+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-2-z\\z=-1\end{matrix}\right.\Leftrightarrow y=z=-1\)

Vậy (x;y;z) = (2;-1;-1) ; (-1 ; 2 ; -1) ; (-1 ; -1 ; 2)

em cảm ơn ạ