Xác định các hệ số a,b,c biết rằng a),(2x_5) (3x+b) = ax^2 +x +c . b), ( ax + b)(x^2_x_1) = ax^3 + cx^2 _ 1.

Đáp án:

 $a)  {\left\{\begin{aligned}a=6\\b=8\\ c=-40\end{aligned}\right.}\\
b)
 {\left\{\begin{aligned}c=2\\ a=-1\\ b=1\end{aligned}\right.}\\$

Giải thích các bước giải:

 $a) (2x-5)(3x+b)=ax^2+x+c\\
\Leftrightarrow 6x^2+2bx-15x-5b=ax^2+x+c\\
\Leftrightarrow 6x^2+(2b-15)x-5b=ax^2+x+c\\
\Rightarrow {\left\{\begin{aligned}a=6\\2b-15=1\\ -5b=c\end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}a=6\\2b=1+15=16\\ -5b=c\end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}a=6\\b=8\\ -5.8=c\end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}a=6\\b=8\\ c=-40\end{aligned}\right.}\\
b)
 (ax+b)(x^2-x-1)=ax^3+cx^2-1\\
\Leftrightarrow ax^3-ax^2-ax+bx^2-bx-b=ax^3+cx^2-1\\
\Leftrightarrow ax^3+(-ax^2+bx^2)+(-ax-bx)-b=ax^3+cx^2-1\\
\Leftrightarrow ax^3+(-a+b)x^2+(-a-b)x-b=ax^3+cx^2-1\\
\Rightarrow {\left\{\begin{aligned}-a+b=c\\ -a-b=0\\ -b=-1\end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}-a+b=c\\ -a-b=0\\ -b=-1\end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}-a+1=c\\ -a-1=0\\ b=1\end{aligned}\right.}\\
\Leftrightarrow  {\left\{\begin{aligned}-(-1)+1=c\\ a=-1\\ b=1\end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}c=2\\ a=-1\\ b=1\end{aligned}\right.}\\$