Sửa đề \(3x^2\left(ax^2-2bx-3c\right)=3x^4-12x^3+27x^2\)

\(\Leftrightarrow3x^2\left(ax^2-2bx-3c\right)=3x^2\left(x^2-4x+9\right)\)

Đồng nhất 2 đa thức ta có:

\(\left\{{}\begin{matrix}ax^2=x^2\\-2bx=-4x\\-3c=9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-2\\c=-3\end{matrix}\right.\)

\(3x^2.\left(ax^2-2bx-3c\right)=3x^4-12x^3+27x^2\)

\(\Leftrightarrow3ax^4-6bx^3-9cx^2=3x^4-12x^3+27x^2\)

\(\Leftrightarrow\hept{\begin{cases}3a=3\\-6b=-12\\-9c=27\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\c=-3\end{cases}}}\)

Vậy a=1;b=2;c=-3

\(3x^2\left(ax^2-2bx-3c\right)=2x^4-12x^3+27x^2\)

\(\Leftrightarrow3ax^4-6bx^3-9cx^2=3\cdot1\cdot x^4-6\cdot2x^3-9\cdot\left(-3\right)x^2\)

\(\Leftrightarrow a=1;b=2;c=-3\)

Vậy \(a=1;b=2;c=-3\)

1. \(3x^2\left(ax^2-2bx-3c\right)=3x^2\left(x^2-4x+27\right)\)

\(\Rightarrow\hept{\begin{cases}a=1\\-2b=-4\\-3c=27\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=2\\c=-9\end{cases}}}\)

2. \(\left(x^2+cx+2\right)\left(ax+b\right)=x^3+x^2-2\)

\(\Rightarrow ax^3+bx^2+acx^2+bcx+2ax+2b=x^3+x^2-2\)

\(\Rightarrow ax^3+\left(b+ac\right)x^2+\left(bc+2a\right)x+2b=x^3+x^2-2\)

\(\Rightarrow\hept{\begin{cases}a=1\\b+ac=1\\2b=-2\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b+ac=1\\b=-1\end{cases}\Rightarrow}\hept{\begin{cases}a=1\\b=-1\\c=2\end{cases}}}\)

Câu còn lại tương tự