Ta có :

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

\(\Leftrightarrow ax^3+\left(b-a\right).x^2-\left(a+b\right).x-b\)

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

\(\Leftrightarrow\hept{\begin{cases}b-a=c\\a+b=0\\b=1\end{cases}}\)

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

Vậy ...

Khai triển VT, ta có: \(VT=ax^3+\left(b+ac\right)x^2+\left(bc+2a\right)x+2b=x^3-x^2+2\)

Đồng nhất hệ số ta có hệ điều kiện:

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

a: =>6x^2+2xb-15x-5b=ax^2+x+c

=>6x^2+x(2b-15)-5b=ax^2+x+c

=>a=6; 2b-15=1; -5b=c

=>a=6; b=8; c=-40

b: =>ax^3-ax^2-ax+bx^2-bx-b=ax^3+cx^2-1

=>x^2(-a+b)+x(-a-b)-b=cx^2-1

=>-b=-1; -a+b=c; -a-b=0

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

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

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

\(\Leftrightarrow ax^{3\:}+\left(ac+b\right)x^2+\left(2a+bc\right)x+2b=x^3+x^2-2\)

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

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

\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)

\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)

Vậy \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)