\(\left(x-1\right)\left(x+1\right)-3x-6=6\)    

\(x^2-1^2-3x-6-6=0\)   

\(x^2-1-3x-12=0\)   

\(x^2-3x-13=0\)   

\(\orbr{\begin{cases}x=\frac{3-\sqrt{61}}{2}\\x=\frac{3+\sqrt{61}}{2}\end{cases}}\)

\(\left(x-1\right)\left(x+1\right)-3x-6=6\)

\(\left(x-1\right)\left(x+1\right)-3x=12\)

\(\left(x-1\right)x-\left(x-1\right)1-\left(1+2\right)x=12\)

\(\left(x-1-1+2\right)x-x-1=12\)

\(\left(x-1-1+2-1\right)x=11\)

\(\left(x-1\right)x=11\)

\(x^2-x=11\)

Đk : x > 4

\(x=4\Rightarrow16-4=11\left(\varnothing\right)\)

\(x\in\varnothing\)

a^4+a^3+a^2+a=a^10

a⁴ + a³ + a² + a

= a³( a + 1 ) + a( a + 1 )

= ( a + 1 )( a³ + a )

= a( a + 1 )( a² + 1 )

Ta có: \(x^4+x^3+x^2-1\)

\(=x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3+x-1\right)\)

x⁴ + x³ + x² - 1

= x³( x + 1 ) + ( x - 1 )( x + 1 )

= ( x + 1 )( x³ + x - 1 )

a) 2x - x³ + 4y - 8y³

= ( 2x + 4y ) - ( x³ + 8y³ )

= 2( x + 2y ) - ( x + 2y )( x² - 2xy + 4y² )

= ( x + 2y )( 2 - x² + 2xy - 4y² )

b) -3x² + 11x + 14

= -3x² + 14x - 3x + 14

= -x( 3x - 14 ) - ( 3x - 14 )

= ( 3x - 14 )( -x - 1 )

Sai đề bạn ơi

3x (5x² - 2x - 1)

Bài 1:

a) \(3x^2-9x=3x\left(x-3\right)\)

b) \(x^2-4x+4=\left(x-2\right)^2\)

c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)

Bài 2: 

a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)

b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)

\(=\left(67+33\right)^2=100^2=10000\)

Bài 3:

\(x\left(x-3\right)+2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

Vậy \(x=-2\)hoặc \(x=3\)

B1:

a) \(3x^2-9x=3x.\left(x-3\right)\)

b) \(x^2-4x+4=\left(x-2\right)^2\)

c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)

B2:

a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)

b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)

B3:

\(x\left(x-3\right)+2\left(x-3\right)=0\)

\(\left(x-3\right).\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

Phân tích hả ?

1) ( x⁴ + 2 )² - 8x⁴

= x⁸ + 4x⁴ + 4 - 8x⁴

= x⁸ - 4x⁴ + 4 = ( x⁴ - 2 )²

2) 9( x + 5 )² - ( x + 7 )²

= 3²( x + 5 )² - ( x + 7 )²

= ( 3x + 15 )² - ( x + 7 )²

= [ ( 3x + 15 ) - ( x + 7 ) ][ ( 3x + 15 ) + ( x + 7 ) ]

= ( 3x + 15 - x - 7 )( 3x + 15 + x + 7 )

= ( 2x + 8 )( 4x + 22 )

= 2( x + 4 )2( 2x + 11 )

= 4( x + 4 )( 2x + 11 )

3) 4a + 4ab⁶ + ab¹²

= a( 4 + 4b⁶ + b¹² )

= a( 2 + b⁶ )²