\(3x^4y-12x^2y^3=3x^2y\left(x^2-4y^2\right)=3x^2y\left(x-2y\right)\left(x+2y\right)\)

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

\(x^3+3x^2+4x+12=x^2\left(x+3\right)+4\left(x+3\right)=\left(x^2+4\right)\left(x+3\right)\)

\(3x^2-6xy+3y^2-27=3\left[\left(x-y\right)^2-9\right]=3\left(x-y-3\right)\left(x-y+3\right)\)

a) \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\left(2x-3y+4xy\right)\)

b) \(3x^2-5x-3xy+5y\)

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

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

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

c) \(5a^3-20a\)

\(=5a\left(a^2-4\right)\)

\(=5a\left(a-2\right)\left(a+2\right)\)

d) \(2x+2y+x^2+2xy+y^2\)

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

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

a) x³+4x-5 = x³-x²+x²+4x-5=(x³-x²)+(x²-x)+(5x-5)=x²(x-1)+x(x-1)+5(x-1)=(x²+x+5)(x-1)

b) x³-3x²+4=x³-2x²-x²+4=(x³-2x²)-(x²-4)=x²(x-2)-(x-2)(x+2)=(x²-x+2)(x-2)

c) x³+2x²+3x+2=x³+x²+x²+x+2x+2=(x³+x²)+(x²+x)+(2x+2)=x²(x+1)+x(x+1)+2(x+1)=(x²+x+2)(x+1)

d) bạn xem lại đề đúng ko

e) (x²+3x)²-2(x²+3x)-8=x⁴+6x³+9x²-2x²-6x-8=x⁴+6x³+7x²-6x-8=x⁴-x³+7x³-7x²+14x²-14x+8x-8=(x⁴-x³)+(7x³-7x²)+(14x²-14x)+(8x-8)=x³(x-1)+7x²(x-1)+14x(x-1)+8(x-1)=(x³+7x²+14x+8)(x-1)=(x³+x²+6x²+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

f) (x²+4x+10)²-7(x²+4x+11)+7=(x²+4x+10)²-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)

a) Ta có: \(x^3+4x-5\)

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

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

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

b) Ta có: \(x^3-3x^2+4\)

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

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

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

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

c) Ta có: \(x^3+2x^2+3x+2\)

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

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

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

d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)

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

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

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

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

a)\(A=3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)b) \(A=\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)

c) \(A=x^2+y^2+2xy+yz+zx=\left(x+y\right)^2+z\left(x+y\right)=\left(x+y\right)\left(x+y+z\right)\)

\(a.3x^2-3y^2-2\left(x-y\right)^2\\ =3\left(x^2-y^2\right)-2\left(x-y\right)^2\\ =3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\\ =\left(x-y\right)\left[3\left(x+y\right)-2.\left(x-y\right)\right]=\left(x-y\right)\left(x+5y\right)\\ b.x^2-y^2-2x-2y\\ =\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\\ =\left(x+y\right)\left(x-y-2\right)\\ c.\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\\ =\left(x-1\right)\left(2x+1\right)\left[1+3\left(x+2\right)\right]\\ =\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\\ d.\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)-\left(5-x\right)\left(2x+1\right)\\ =\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)\\ =\left(x-5\right)\left[\left(x-5\right)+\left(x+5\right)+\left(2x+1\right)\right]\\ =\left(x-5\right)\left(4x+1\right)\)

a) 3x²-3y²-2(x-y)²

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

1) \(x^2-y^2-2x-2y\)

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

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

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

2) \(3x^2-3y^2-2\left(x-y\right)^2\)

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

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

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

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

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

1) x² - y² - 2x - 2y

= (x² - y²) - (2x + 2y)

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

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

2) 3x² - 3y² - 2(x - y)²

= (3x² - 3y²) - 2(x - y)²

= 3(x² - y²) - 2(x - y)²

= 3(x - y)(x + y) - 2(x - y)²

= (x - y)[3(x + y) - 2(x - y)]

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

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

a) x² ( x+ 2y) -x -2y

= x² ( x+ 2y) -(x+2y)

= (x²-1)(x+2y)

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

b)3x²- 3y² -2 (x-y)²

= 3(x²-y²) -2 (x-y)²

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

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

c) x²- 2x-4y² - 4y

= (x²-4y²)-(2x+4y)

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

d) x³ - 4x² - 9x +36

= (x³+3x²)-(7x²+21x)+(12x+36)

= x²(x+3)-7x(x+3)+12(x+3)

=(x²-7x+12)(x+3)

\(=\left[\left(x^2-3x\right)-\left(4x-12\right)\right]\left(x+3\right)\\ =\left[x\left(x-3\right)-4\left(x-3\right)\right]\left(x+3\right)=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)

cảm ơn bạn nhiều nha!