Dùng pp hệ số bất định nha bạn

Đặt

A = \(x^4-4x^3-7x^2+16x-3\)

A \(=\left(x^2+ax+1\right)\left(x^2+bx-3\right)\)

A \(=x^4+ax^3+x^2+bx^3+abx^2+bx-3x^2-3ax-3\)

A \(=x^4+\left(a+b\right)x^3+\left(1+ab-3\right)x^2+\left(b-3a\right)x-3\)

Đồng nhất 2 đa thức ta được :

\(\left\{{}\begin{matrix}a+b=-4\\-2+ab=-7\\b-3a=16\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=-5\\b=1\end{matrix}\right.\)

\(\Rightarrow A=\left(x^2-5x+1\right)\left(x^2+x-3\right)\)

Đặt \(M=\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\)

Xét \(M\times\dfrac{c}{a-b}=1+\dfrac{c}{a-b}\left(\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{c}{a-b}\left(\dfrac{b^2-bc+ac-a^2}{ab}\right)=1+\dfrac{c}{a-b}\left(\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab}\right)=1+\dfrac{c}{a-b}\left(\dfrac{\left(b-a\right)\left(a+b-c\right)}{ab}\right)\)

Vì \(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow a+b-c=-2c\)

\(\Rightarrow M\times\dfrac{c}{a-b}=1+\dfrac{c}{a-b}\times\dfrac{\left(a-b\right)2c}{ab}=1+\dfrac{2c^2}{ac}=1+\dfrac{2c^3}{abc}\)

Tương tự \(M\times\dfrac{a}{b-c}=1+\dfrac{2a^3}{abc}\)

\(M\times\dfrac{b}{c-a}=1+\dfrac{2b^3}{abc}\)

\(\Rightarrow A=3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}\)

Mà do \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)

\(\Rightarrow A=9\)