ta co' 
(x+a).(x-4)-7=(x+b).(x+c) 
nen voi x=4 thi 
-7=(4+b)(4+c)=-7.1=7.(-1) 
do a,c,b∈Z va b,c co vai tro nhu nhau nen gia su b>=c 
co 2 TH xay ra 
**{4+b=7│4+c=-1}↔{b=3│c=-5}suy ra a=2 
ta co(x+2)(x-4_-7=(x+3)(x-5) 
** {4+b=1│4+c=-7}↔{b=-3│c=-11} suy ra a=-10 
ta co(x-10)(x-4)-7=(x-3)(x-11)

1. \(M=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1\)

\(=\left[\left(a+1\right)\left(a+4\right)\right]\left[\left(a+2\right)\left(a+3\right)\right]+1\)

\(=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)

\(=\left(a^2+5a+4\right)^2+2\left(a^2+5a+4\right)+1\)

\(=\left(a^2+5a+5\right)^2\) 

=> Đpcm

M = ( a + 1 )( a + 2 )( a + 3 )( a + 4 ) + 1

    = [ ( a + 1 )( a + 4 ) ][ ( a + 2 )( a + 3 ) ] + 1

    = [ a² + 5a + 4 ][ a² + 5a + 6 ] + 1

Đặt t = a² + 5a + 4

M <=> t[ t + 2 ] + 1

      = t² + 2t + 1

      = ( t + 1 )²

      = ( a² + 5a + 4 + 1 )² = ( a² + 5a + 5 )² ( đpcm )

( x² + x + 1 )( x² + x + 2 ) - 12 (*)

Đặt t = x² + x + 1

(*) <=> t( t + 1 ) - 12

       = t² + t - 12

       = t² - 3t + 4t - 12

       = t( t - 3 ) + 4( t - 3 )

       = ( t - 3 )( t + 4 )

       = ( x² + x + 1 - 3 )( x² + x + 1 + 4 )

       = ( x² + x - 2 )( x² + x + 5 )

       = ( x² + 2x - x - 2 )( x² + x + 5 )

       = [ x( x + 2 ) - 1( x + 2 ) ]( x² + x + 5 )

       = ( x + 2 )( x - 1 )( x² + x + 5 )

(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

\(x^3-9x^2+26x-24\)

\(=x^3-4x^2-5x^2+20x+6x-24\)

\(=\left(x-4\right)\left(x^2-5x+6\right)\)

\(=\left(x-4\right)\left(x-2\right)\left(x-3\right)\)

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