Lấy: $(x+1)(x+2).PT1-PT2$ Ta có: $(y-2)(2x^3+x^2+x+y)=0$$y=2,2x^3+x^2+x+y=0$\begin{cases}2x^3+x^2+x+y=0(*) \\ x^2-2xy+x+y=0(**) \end{cases}Lấy (*)-(**):$2x^3+2xy=0<=>2x(x^2+y)=0$xong game nhé
Bài 8Lấy: $(x+1)(x+2).
(1
)-
(2
)$ Ta có:
$(y-2)(2x^3+x^2+x+y)=0$
$
\Leftrightarrow \left[\ \begin{array}{l} y=2
\\ 2x^3+x^2+x+y=0
\end{array} \right.$Xét hệ: $\begin{cases}2x^3+x^2+x+y=0(*) \\ x^2-2xy+x+y=0(**) \end{cases}
$Lấy
$(*)-(**)
\Leftrightarrow 2x^3+2xy=0
\Left
rig
ht
arrow 2x(x^2+y)=0$xong game nhé
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