\(\left(x^2-2x+3\right)\left(2x-x^2+6\right)=18\)
\(\Leftrightarrow\left(x^2-2x+3\right)\left[-\left(x^2-2x+3\right)+9\right]=18\)
Đặt \(t=x^2-2x+3\left(t\ge0\right)\) ta có:
\(t\left(-t+9\right)=18\)
\(\Leftrightarrow t^2-9t+18=0\)
\(\Leftrightarrow\left(t-6\right)\left(t-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-6=0\\t-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=6\\t=3\end{matrix}\right.\) (tmđk)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+3=6\\x^2-2x+3=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-2x-3=0\\x^2-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)\left(x-1\right)=0\\x\left(x-2\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\\x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=0\\x=2\end{matrix}\right.\)
Vậy phương trình đã cho có \(S=\left\{3;1;0;2\right\}\)
2x -x2 +6 = -(x2 -2x +3) +9
đặt t = x2 -2x+3, nam võ có: -t(t+9) =18
máy tinh giúp a tiếp
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