a)ta có:

\(f\left(x\right):\left(x+1\right)\: dư\: 6\Rightarrow f\left(x\right)-6⋮\left(x+1\right)\\ hay\: 1-a+b-6=0\\ \Leftrightarrow b-a-5=0\Leftrightarrow b-a=5\left(1\right)\)

tương tự: \(2^2+2a+b-3=0\\ 2a+b=-1\left(2\right)\)

từ (1) và(2) => \(\left\{{}\begin{matrix}b-a=5\\2a+b=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=-2\\b=3\end{matrix}\right.\)

Câu a :

Theo đề bài ta có hệ phương trình :

\(\left\{{}\begin{matrix}f\left(-1\right)=1-a+b=6\\f\left(2\right)=4+2a+b=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-a+b=5\\2a+b=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=-2\\b=3\end{matrix}\right.\)

Vậy đa thức \(f\left(x\right)=x^2-2x+3\)

\(f\left(x\right)⋮g\left(x\right)\)

\(\Rightarrow\left\{{}\begin{matrix}1^4-3.1^3+b.1^2+a.1+b=0\\\left(-1\right)^4-3.\left(-1\right)^3+b.\left(-1\right)^2+a.\left(-1\right)+b=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2b+a=2\\2b-a=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3\\b=-\dfrac{1}{2}\end{matrix}\right.\)

Gọi r là số dư 

Ta có: A(x)=B(x).(x+1)+r

          A(x)=C(x).(x-1)+r

=> A(1)=a+b+c=C(x).0+r=> a+b+c=r     (1)

    A(-1)=a-b+c =B(x).0+r=> a-b+c=r       (2)

lẤY (1)-(2) ta có: 2b=0=> b=0

Vì \(f\left(x\right)⋮x-2;f\left(x\right):x^2-1\) dư 1\(\Rightarrow\left\{{}\begin{matrix}f\left(x\right)=g\left(x\right)\cdot\left(x-2\right)\\f\left(x\right)=q\left(x\right)\left(x^2-1\right)+x=q\left(x\right)\left(x-1\right)\left(x+1\right)+x\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=0\\f\left(1\right)=1\\f\left(-1\right)=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}32+4a+2b+c=0\\2+a+b+c=1\\2+a-b+c=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4a+2b+c=-32\left(1\right)\\a+b+c=-1\left(2\right)\\a-b+c=-3\left(3\right)\end{matrix}\right.\)

 Trừ từng vế của (2) cho (3) ta được:

\(\Rightarrow2b=2\Rightarrow b=1\)

Thay b=1 vào lần lượt (1) ,(2),(3) ta được:

\(\Rightarrow\left\{{}\begin{matrix}4a+2+c=-32\\a+1+c=-1\\a-1+c=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+c=-34\\a+c=-2\\a+c=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4a+c=-34\left(4\right)\\a+c=-2\left(5\right)\end{matrix}\right.\)

Trừ từng vế của (4) cho (5) ta được:

\(\Rightarrow3a=-32\Rightarrow a=-\dfrac{32}{3}\Rightarrow c=-2+\dfrac{32}{3}=\dfrac{26}{3}\) Vậy...

Ta có: \(f\left(x\right)=ax^2+bx+c\)

\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=a\cdot2^2+2b+c=4a+2b+c\\f\left(-5\right)=a\cdot\left(-5\right)^2-5b+c=25a-5b+c\end{matrix}\right.\)

\(\Rightarrow f\left(2\right)\cdot f\left(-5\right)=\left(4a+2b+c\right)\left(25a-5b+c\right)\)

Lại có:\(25a-5b+c=29a+2c-c-4a-5b\)

\(=3b-c-4a-5b=-2b-c-4a=-\left(4a+2b+c\right)\)

\(\Rightarrow f\left(2\right)\cdot f\left(-5\right)=-\left(4a+2b+c\right)\left(4a+2b+c\right)\)

\(=-\left(4a+2b+c\right)^2\le0\forall a,b,c\)

=> Q(2)=a2^2+2b+c=4a+2b+c

Q(-1)=a(-1)^2+(-1)b+c=a-b+c

Ta có: 4a+2b+c=5a+b+2c-a+b-c=0-a+b-c=-a+b-c

=>Q(2).Q(-1)=(4a+2b+c).(a-b+c)=(-a+b-c).(a-b+c)=-(a-b+c).(a-b+c)≤ 0 với mọi a,b,c

Nhầm đây mới là câu trả lời:

Ta có:Q(x)=ax²+bx+x

=>Q(2)=a2^2+2b+c=4a+2b+c

Q(-1)=a(-1)^2+(-1)b+c=a-b+c

Ta có: 4a+2b+c=5a+b+2c-a+b-c=0-a+b-c=-a+b-c

=>Q(2).Q(-1)=(4a+2b+c).(a-b+c)=(-a+b-c).(a-b+c)=-(a-b+c).(a-b+c)≤ 0 với mọi a,b,c

\(f\left(x\right)=ax^2+bx+c\)

=> \(f\left(-2\right)=4a-2b+c=-3\)

Có f(x) chia cho x và x + 4 đều dư 5

=> \(\left\{{}\begin{matrix}f\left(0\right)=0+c=5\\f\left(-4\right)=16a-4b+c=5\end{matrix}\right.\)

Ta có hpt:

\(\left\{{}\begin{matrix}4a-2b+c=-3\\c=5\\16a-4b+c=5\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}c=5\\2\left(2a-b\right)=-8\\4\left(4a-b\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}c=5\\b=4a\\2a-b=-4\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=8\\c=5\end{matrix}\right.\)

Khi đó \(f\left(x\right)=2x^2+8x+5\)

- Để đa thức f(x) trên là một đa thức không thì :

\(\left\{{}\begin{matrix}2m-n+1=0\\m-3n+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m-n=-1\\m-3n=-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m=-\dfrac{1}{5}\\n=\dfrac{3}{5}\end{matrix}\right.\)

Vậy ...