Question

Xác định a,b,c biết:

a) (ax\(^2\)+bx+c)(x+1)=x\(^3\)+8x\(^2\)+19x+12

b) (ax\(^2\)+bx+c)(x+3)=x\(^3\)+2x\(^2\)-3x

c) (x\(^2\)+cx+2)(ax+b)=x\(^2\)+x\(^2\)-2

d) (x\(^2\)+cx+1)(ax+b)=x\(^3\)-3x+2

Answer 1

\(\left(ax^2+bx+c\right)\left(x+1\right)=ax^3+\left(a+b\right)x^2+\left(b+c\right)x+c\)

đồng nhất đa thức trên với đa thức đã cho ta được

\(\left\{{}\begin{matrix}a=1\\a+b=8\\b+c=19\\c=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=7\\c=12\end{matrix}\right.\)

3 phần kia làm tương tự

Answer 2

b: \(\left(ax^2+bx+c\right)\left(x+3\right)\)

\(=ax^3+3ax^2+bx^2+3bx+cx+3c\)

\(=ax^3+x^2\left(3a+b\right)+x\left(3b+c\right)+3c\)

Theo đề, ta có:

\(\left\{{}\begin{matrix}3c=0\\3b+c=-3\\3a+b=2\\a=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=0\\b=-1\\a=1\end{matrix}\right.\)

c: \(\left(x^2+cx+2\right)\left(ax+b\right)\)

\(=a\cdot x^3+bx^2+ac\cdot x^2+bc\cdot x+2a\cdot x+2b\)

\(=a\cdot x^3+x^2\left(b+ac\right)+x\left(bc+2a\right)+2b\)

Theo đề, ta có: 2b=-2; bc+2a=0; b+ac=1; a=1

=>b=-1; a=1; c=2

d: \(\left(x^2+cx+1\right)\left(ax+b\right)\)

\(=a\cdot x^3+bx^2+ac\cdot x^2+bc\cdot x+a\cdot x+b\)

\(=a\cdot x^3+x^2\left(b+ac\right)+x\left(bc+a\right)+b\)

Theo đề, ta có:

b=2; bc+a=-3; b+ac=0; a=1

=>b=2; a=1; bc=-3-a=-3-1=-4

=>b=2; a=1; 2c=-4

=>b=2; a=1; c=-2