Ta có:
\(4A+3B-\left(4A+2B\right)=x^2-2x+1-\left(x^2-2x+9\right)\)
\(\Rightarrow B=-8\)
Thay B vào \(4A+2B=x^2-2x+9\) được:
\(4A+2.\left(-8\right)=x^2-2x+9\)
\(\Rightarrow4A=x^2-2x+9+16\)
\(\Rightarrow4A=x^2-2x+25\)
\(\Rightarrow A=\dfrac{x^2-2x+25}{4}\)
Vậy...
\(\left\{{}\begin{matrix}4A+2B=x^2-2x+9\\4A+3B=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A+3B=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A+3.\left(-8\right)=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A-24=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A=x^2-2x+25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\A=\dfrac{x^2-2x+25}{4}\end{matrix}\right.\)
