a) \(x^2-xy+x-y\)

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

\(=\left(x+1\right)\left(x-y\right)\)

b) \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

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

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

\(2xy-x^2-y^2+16\)

\(=16-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)

`#3107.101107`

`(4x - 1)^2 - 121`

`= (4x - 1)^2 - (11)^2`

`= (4x - 1 - 11)(4x - 1 + 11)`

`= (4x - 12)(4x + 10)`

`= 4(x - 3) * 2(2x + 5)`

`= 8(x - 3)(2x + 5)`

_____

`x^6 - y^6`

`= (x^3)^2 - (y^3)^2`

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

`= (x - y)(x^2 + xy + y^2)(x + y)(x^2 - xy + y^2)`

`= (x - y)(x + y)(x^2 + xy + y^2)`

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Sử dụng các HĐT:

`@` `A^2 - B^2 = (A - B)(A + B)`

`@` `A^3 - B^3 = (A - B)(A^2 + AB + B^2)`

`@` `A^3 + B^3 = (A + B)(A^2 - AB + B^2).`

a: \(\left(4x-1\right)^2-121\)

\(=\left(4x-1\right)^2-11^2\)

\(=\left(4x-1-11\right)\left(4x-1+11\right)\)

\(=\left(4x-12\right)\left(4x+10\right)\)

\(=8\left(x-3\right)\left(2x+5\right)\)

b: \(x^6-y^6\)

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

\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(x^6-y^6\)

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

\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)

x^{6 -} y⁶

= ( x³ )² - ( y³ ) ²

⁼ ( x³ + y³) ( x³ - y³ )

= ( x - y ) ( x² + xy + y^{2 )}.( x + y ) ( x² - xy + y^{2 )}

\(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3=\left(x^2+y^2\right).\left(x^4-x^2y^2+y^4\right)\\ ---\\ 0,04-9x^2=\left(0,2\right)^2-\left(3x\right)^2=\left(0,2-3x\right)\left(0,2+3x\right)\\ ---\\ 32x^2-2\left(y-1\right)^2=2\left[16x^2-\left(y-1\right)^2\right]=2\left[\left(4x\right)^2-\left(y-1\right)^2\right]\\ =2\left(4x-y+1\right)\left(4x+y-1\right)\)

a) x⁴ - y⁴

= (x²)² - (y²)²

= (x² - y²)(x² + y²)

= (x - y)(x + y)(x² + y²)

b) 1 - 8x³y⁶

= 1³ - (2xy²)³

= (1 - 2xy²)(1 + 2xy² + 4x²y⁴)

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

\(x^6-y^6\)

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

\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)

x⁶y - 5x⁵ - 4x⁴y + 20x³ 

= ( x⁶y - 5x⁵ ) - ( 4x⁴y - 20x³ ) 

= x⁵( xy - 5 ) - 4x³( xy - 5 )

= ( x⁵ - 4x³ )( xy - 5 ) = x³( x² - 4 )( xy - 5 ) 

= x³( x - 2 )(x + 2 )( xy - 5 )

 = x^3.(x^3y-5x^2-4xy+20)

 = x^3.[(x^3y-5x^2)-(4xy-20)]

 = x^3.(y-5).(x^2-4) = x^3.(x-2).(x+2).(y-5)

k mk nha

1. 

\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)

2. 

\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt

Bài 1:

Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)

\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)

\(=6x\left(2x+1\right)\cdot2y\)

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

Bài 2: 

Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

\(3x\cdot\left(x-y\right)^2-6\cdot\left(y-x\right)\)

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

\(=\left(x-y\right)\left[3x\left(x-y\right)+6\right]\)

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

(x - y)(3x² - 3xy + 6)