(4x-x-y)(4x+x+y)

= (3x-y)(5x+y)

sorry phải là

(2x-x-y)(2x+x+y)

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

\(4x^2-\left(x+y\right)^2\)

\(=\left(2x-x-y\right)\left(2x+x+y\right)\)   ( dùng hằng đẳng thức hiệu 2 bình phương)

$x(x+y)+4x+4y$

$=x(x+y)+4(x+y)$

$=(x+y)(x+4)$

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

\(a^2-7a+12=\left(a^2-3a\right)-\left(4a-12\right)=a\left(a-3\right)-4\left(a-3\right)=\left(a-4\right)\left(a-3\right)\) 

\(4a^2+4a-3=4a^2-2a+\left(6a-3\right)=2a\left(2a-1\right)+3\left(2a-1\right)=\left(2a+3\right)\left(2a-1\right)\)

X² - 6x + 5

= x² - 6x + 5 + 4 - 4 

= x² - 6x + 9 - 2²

= ( x - 3 )² - 2² 

= ( x - 3 - 2 ) ( x - 3 + 2 ) 

X² - 7x + 12

= X² - 7x + 12 + 0.25 - 0,25

= X² - 7x + 12,25 - ( 0,5 )²

= ( X - 3,5 )² -  ( 0,5 )²

= ( X - 3,5 - 0,5 ) ( X - 3,5 + 0,5 )

Ta có: 4x² - y² + 4x + 4y - 3

= (4x² - 4x + 1) - (y² - 4y + 4)

= (2x - 1)² - (y - 2)²

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

= (2x - y + 1)(2x + y - 3)

\(4x^2-y^2+4x+4y-3\)

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

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

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

\(=\left(2x+y-1\right)\left(2x-y+3\right)\)

Ta có : 4x² - y² + 4x + 4y - 3

         = 4x² - y² + 4x + 4y + 1 - 4 

        = ( 4x² + 4x + 1 ) - ( y² - 4y + 4 )

       = ( 2x + 1 )² - ( y - 2 )²

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

     = ( 2x - y + 3 )( 2x + y - 1 )

\(x\left(x+1\right)\left(x^2+x-5\right)-6\)

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

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

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

\(=\left(x^2+x\right)\left(x^2+x+1\right)-6\left(x^2+x+1\right)\)

\(=\left(x^2+x-6\right)\left(x^2+x+1\right)\)

\(=\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)

\(x^4+x^2-2=x^4-x^2+2x^2-2 \)

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

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

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

a: \(2x^3+x^2-13x+6\)

\(=2x^3-4x^2+5x^2-10x-3x+6\)

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

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

\(=\left(x-2\right)\left(x+3\right)\left(2x-1\right)\)

b: \(2x^2+y^2-6x+2xy-2y+5=0\)

\(\Leftrightarrow x^2+2xy+y^2+x^2-4x+4-2x-2y+1=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-2\right)^2-2\left(x+y\right)+1=0\)

\(\Leftrightarrow\left(x-2\right)^2+\left(x+y-1\right)^2=0\)

=>x-2=0 và x+y-1=0

=>x=2 và y=-1

Ta có : 5x³ - 45x

= 5x(x² - 9)

= 5x(x - 3)(x + 3)

b ) Ta có : 3x² - 7x - 6 

= 3x² - 9x + 2x - 6 

= 3x (x - 3) + 2(x - 3)

= (x - 3)(3x + 2)

\(-3x^2+4x-2020\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{2020}{3}\right)\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{4}{9}+\frac{6056}{9}\right)\)

\(=-3\left[\left(x-\frac{2}{3}\right)^2+\frac{6056}{9}\right]\)

\(=-3\left(x-\frac{2}{3}\right)^2-\frac{6056}{3}\ge-\frac{6056}{3}\)

(Dấu "=" \(\Leftrightarrow x-\frac{2}{3}=0\Leftrightarrow x=\frac{2}{3}\))