\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

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

\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)

Bài 1:

\(x^2-6x+9-y^2\)

\(=\left(x-3\right)^2-y^2\)

\(=\left(x-3+y\right)\left(x-3-y\right)\)

Bài 2:

\(x^2-x-12=0\)

\(\Leftrightarrow x^2-4x+3x-12=0\)

\(\Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\\x-4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)

1. x²+6x-9-y²

=-(x²-6x+y²)-3²

=-(x-y)²-3²

=(-x+y-3)(-x+y+3)

2. x² - x - 12 = 0

=> x² - 4x + 3x - 12 = 0

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

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

\(\Rightarrow\left[\begin{array}{nghiempt}x+3=0\\x-4=0 \end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)

Vậy x=-3; x=4

x¹⁰+x⁵+1

= x¹⁰+x⁹+x⁸-x⁹-x⁸-x⁷+x⁷+x⁶+x⁵-x⁶-x⁵-x⁴+x⁵+x⁴+x³-x³-x²-x+x²+x+1

= x⁸(x²+x+1)-x⁷(x²+x+1)+x⁵(x²+x+1)-x⁴(x²+x+1)+x³(x²+x+1)-x(x²+x+1)+(x²+x+1)

= (x²+x+1)(x⁸-x⁷+x⁵-x⁴+x³-x+1)

x(x+2)(x^2+2x+2)+1 = (x^2+2x)(x^2+2x+1)+1

Đặt x^2+2x+1=y ta được:

(y-)(y+1)+1=y^2-1+1=y^2

= (x^2+2x+1)^2

= ( x + 1 )^4

\(\left(1+2x\right).\left(1-2x\right)-x.\left(x+2\right).\left(x-2\right)\))

\(=1-\left(2x\right)^2-x.x^2-2^2\)

\(=1-4x^2-x^3-4\)

Ko bt có đúng ko nữa

( 1 + 2x ) ( 1 - 2x ) - x ( x + 2 ) ( x - 2 ) 

= 1 - 4x² - x ( x² - 4 )

= 1 - 4x² - x³ + 4x

= - ( x³ + 4x² - 4x - 1 )

= - ( x³ - x² + 5x² - 5x + x - 1 )

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

= - ( x - 1 ) ( x² + 5x + 1 )