\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

\(4x^2-y^2+8x-16\)

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

4x² - y² + 8y - 16

= 4x² - (y² - 8y + 16)

= (2x)² - (y - 4)²

= [2x - (y - 4)][2x + (y - 4)]

= (2x - y +4)(2x + y - 4)

Đề sai.

5x² + 10xy - 4x - 8y

= 5x(x + 2y) - 2(x - 2y)

= (5x - 2)(x + 2y)

=(5x² + 10xy) - (4x + 8y) = 5x.(x + 2y) - 4.(x + 2y) = (5x - 4).(x + 2y)

\(5x^2+10xy-4x-8y=\left(5x^2+10xy\right)-\left(4x+8y\right)=5x\left(x+2y\right)-4\left(x+2y\right)=\left(5x-4\right)\left(x+2y\right)\)

D = x² - 4x - y² - 8y - 12 

= (x² - 4x + 4) - (y² + 8y + 16) 

= (x - 2)² - (y + 4)²

= (x + y + 2)(x - y - 6) 

\(D=x^2-4x-y^2-8y-12\)

\(=x^2-4x-y^2-8y+4-16\)

\(=\left(x^2-4x+4\right)-\left(y^2+8y+16\right)\)

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

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

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

a) \(x^2-3x=x\left(x-3\right)\)

b) \(10x\left(x-y\right)-8y\left(x-y\right)=2\left(x-y\right)\left(5x-4y\right)\)

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

a) x²-3x=x(x-3)

c) x²-9=x²-3²=(x+3)(x-3)

a: \(x^2-3x=\left(x-3\right)\cdot x\)

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

Theo bài ra , ta có : 

4x² - 4xy + 8y² 

= (2x)² - 2.2xy + y² + 7y² 

= (2x + y)² + 7y² 

Chúc bạn hôc tốt =))

\(4x^2-4x-y^2-8y-15\)

\(=\left(4x^2-4x+1\right)-\left(y^2+8y+16\right)\)

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

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

\(4x^2-4x-y^2-8y-15\)

\(=\left(4x^2-4x-1\right)-\left(y^2-8y-16\right)\)

\(=\left(4x^2-4x-1\right)-\left(y^2+8y+16\right)\)

\(=\left[\left(2x\right)^2-4x-1\right]-\left[y\left(y+8\right)\right]+16\)

\(=\left(2x^2-6x\right)\left(y+12\right)\left(y-4\right)\)

...

ghytyyj