\(\dfrac{1}{3}\)

a. ( 6x + 1 ) ² + (6x - 1 )² - 2(1+ 6x ) (6x-1 )

= (6x+1)²-2(6x+1)(6x-1)+(6x-1)²

=[(6x+1)-(6x-1)]²

= (6x+1-6x+1)²

= 2²

=4

e, (x+1)(x+3)(x+5)(x+7)-9

= [(x+1)(x+7)][(x+3)(x+5)]-9

= (x²+7x+x+7)(x²+5x+3x+15)-9

= (x²+8x+7)(x²+8x+15)-9

đặt x²+8x+7=a

= a(a+8)-9

=a²+8a-9

=x²+9x-x-9

=x(x+9)-1(x+9)

=(x-1)(x+9)

thay a=x²+8x+7 vào ta đc

(x2+8x+7-1)(x2+8x+7+9)

= (x2+8x+6)(x2+8x+16)

b; 2x³+x²-4x-12

= 2x³-4x²+5x²-10x+6x-12

=(2x3-4x2)+(5x2-10x)+(6x-12)

=2x²(x-2)+ 5x(x-2)+6(x-2)

= (x-2)(2x²+5x

x + 11 $⋮$ x + 1

vì \(x+1⋮x+1\)

=>\(\left(x+11\right)-\left(x+1\right)⋮\left(x+1\right)\)

=>\(\left(x+11-0x-1\right)⋮\left(x+1\right)\)

=> \(10⋮x+1\)

=>\(x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

ta có bảng sau:

vậy\(x\in\left\{-11;-6;-3;-2;0;1;4;9\right\}\)

a) x² + 4y² + 4x - 8y - 4xy

= (x²-4xy+4y²)+(4x-8y)

= (x²-2y)²+4(x-2y)

= (x-2y)(x²-2y+4)

x³ - x² - 8x + 12 = 0

=>x³+3x²-4x²-12x+4x+12=0

=> (x³+3x²)-(4x²+12x)+(4x+12)=0

=> x²(x+3)-4x(x+3)+4(x+3)=0

=> (x+3)(x2-4x+4)=0

=> (x+3)(x-2)²=0

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

vậy x\(\in\left\{-3;2\right\}\)

(x+1)(x²-x+1)-(x-1)(x²+x+1)

= x³+1-(x³-1)

= x³+1-x³+1

= 2