(x+m)(x-3)+7=(x+a)(x+b)

=>\(x^2-3x+mx-3m+7=x^2+\left(a+b\right)x+ab\)

=>\(x^2+x\left(m-3\right)-3m+7=x^2+\left(a+b\right)x+ab\)

=>m-3=a+b và -3m+7=ab

=>3m-9=3a+3b và -3m+7=ab

=>3m-9-3m+7=3a+3b+ab

=>ab+3a+3b=-2

=>a(b+3)+3b+9=-2+9

=>(a+3)(b+3)=7

=>(a+3;b+3)∈{(1;7);(7;1);(-1;-7);(-7;-1)}

=>(a;b)∈{(-2;4);(4;-2);(-4;-10);(-10;-4)}

TH1: (a;b)∈{(-2;4);(4;-2)}

(x+a)(x+b)

=(x-2)(x+4)

\(=x^2+2x-8\)

(x+m)(x-3)+7=(x-2)(x+4)

=>\(x^2-3x+mx-3m+7=x^2+2x-8\)

=>x(m-3)-3m+7=2x-8

=>m-3=2 và -3m+7=-8

=>m=5

TH2: \(\left(a;b\right)\in\left\lbrace\left(-4;-10\right);\left(-10;-4\right)\right\rbrace\)

=>(x+a)(x+b)=(x-4)(x-10)

(x+m)(x-3)+7=(x-4)(x-10)

=>\(x^2-3x+mx-3m+7=x^2-14x+40\)

=>x(m-3)-3m+7=-14x+40

=>m-3=-14 và -3m+7=40

=>m=-11

x^2+(a-5)x-5a+2=x^2+(b+c)x+bc

=> \(\hept{\begin{cases}a-5=b+c\\2-5a=bc\end{cases}\Leftrightarrow}\)

giải pt nghiệm nguyên=>

a=-2

b=-4

c=-3

c) Cách 1:

x^4+3x^3-x^2+ax+b x^2+2x-3 x^2+x x^4+2x^3-3x^2 - x^3+2x^2+ax+b x^3+2x^2-3x - (a+3)x+b

Để \(P\left(x\right)⋮Q\left(x\right)\)

\(\Leftrightarrow\left(a+3\right)x+b=0\)

\(\Leftrightarrow\hept{\begin{cases}a+3=0\\b=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=-3\\b=0\end{cases}}\)

Vậy a=-3 và b=0 để \(P\left(x\right)⋮Q\left(x\right)\)

a) 

  2n^2-n+2 2n+1 n-1 2x^2+n - -2n+2 -2n-1 - 3

Để \(2n^2-n+2⋮2n+1\)

\(\Leftrightarrow3⋮2n+1\)

\(\Leftrightarrow2n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow n\in\left\{0;1;-2;-1\right\}\)

Vậy \(n\in\left\{0;1;-2;-1\right\}\)để \(2n^2-n+2⋮2n+1\)