Question

[Trang Phương]

Tìm A và B

`4A+2B``=``x^2-2x+9`

`4A+3B``=``x^2-2x+1`

Answer 1

[Nguyễn Tuấn Tú]

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...

Answer 2

[Vũ Đức Thịnh]

\(\left\{{}\begin{matrix}4A+2B=x^2-2x+9\\4A+3B=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A+3B=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A+3.\left(-8\right)=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A-24=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A=x^2-2x+25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\A=\dfrac{x^2-2x+25}{4}\end{matrix}\right.\)