a: \(=x^2\left(2x+3\right)+\left(2x+3\right)\)

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

b: \(=\left(x-4\right)\left(x+3\right)\)

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

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

b) \(=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)

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

d) \(=4xy\left(y-3x+2\right)\)

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

f) \(=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-4z^2\right]=x\left(x+y-2z\right)\left(x+y+2z\right)\)

g) \(=x\left(x^2-2xy+y^2-25\right)=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\)

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

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

a: Ta có: \(x^2-4y^2-2x-4y\)

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

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

c: Ta có: \(x^3+2x^2y-x-2y\)

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

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

d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)

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

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

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

e: Ta có: \(x^3-4x^2-9x+36\)

\(=x^2\left(x-4\right)-9\left(x-4\right)\)

\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)

f: Ta có: \(x^2-y^2-2x-2y\)

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

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

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

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

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

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

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

c) \(9x^2-3x+2y-4y^2\)

\(=9x^2-4y^2-3x+2y\)

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

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

d) \(4x^2-4xy+2x-y+y^2\)

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

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

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

e) \(x^3+3x^2+3x+1-y^3\)

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

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

g) \(x^3-2x^2y+xy^2-4x\)

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

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

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

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

a) (x + 2y)² - (x - y)²

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

= 3y(2x + y)

b) (x + 1)³ + (x - 1)³

= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]

= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)

= 2x(x² + 3)

c) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

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

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

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

e) x³ + 3x² + 3x + 1 - y³

= (x³ + 3x² + 3x + 1) - y³

= (x + 1)³ - y³

= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]

= (x - y + 1)(x² + 2x + 1 + xy + y + y²)

g) x³ - 2x²y + xy² - 4x

= x(x² - 2xy + y² - 4)

= x[(x² - 2xy + y²) - 4]

= x[(x - y)² - 2²]

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

a) x3 + 2x2y + xy2 – 4x = x(x2 + 2xy + y2– 4) = x[(x+y)2-4]

= x(x + y + 2)(x + y – 2)

Bạn thử xem lại đề câu d nhé.

a) Ta có: \(4x\left(2x-3y\right)-8y\left(3y-2x\right)\)

\(=4x\left(2x-3y\right)+8y\left(2x-3y\right)\)

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

b) Ta có: \(4x^2-4xy+y^2-9z^2\)

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

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

c) Ta có: \(x^2y+yz+xy^2+xz\)

\(=xy\left(x+y\right)+z\left(x+y\right)\)

\(=\left(x+y\right)\left(xy+z\right)\)

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

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

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

2x2 + 4x + 2 – 2y2 (có nhân tử chung là 2)

= 2.(x2 + 2x + 1 – y2) (Xuất hiện x2 + 2x + 1 là hằng đẳng thức)

= 2[(x2 + 2x + 1) – y2]

= 2[(x + 1)2 – y2] (Xuất hiện hằng đẳng thức (3))

= 2(x + 1 – y)(x + 1 + y)

= x(x^2 + 2xy + y^2 - 25z^2)

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

a: \(50x^5-8x^3\)

\(=2x^3\left(25x^2-4\right)\)

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

b: \(x^4-5x^2-4y^2+10y\)

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

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

c: \(36a^2+12a+1-b^2\)

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

\(=\left(6a+1-b\right)\left(6a+1+b\right)\)

d: \(x^3+y^3-xy^2-x^2y\)

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

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

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

e: Ta có: \(4x^2+4x-3\)

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

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

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

f: Ta có: \(9x^4+16x^2-4\)

\(=9x^4+18x^2-2x^2-4\)

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

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

g: Ta có: \(-6x^2+5xy+4y^2\)

\(=-6x^2+8xy-3xy+4y^2\)

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

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

h: Ta có: \(\left(x^2+4x\right)^2+8\left(x^2+4x\right)+15\)

\(=\left(x^2+4x\right)^2+3\left(x^2+4x\right)+5\left(x^2+4x\right)+15\)

\(=\left(x^2+4x+3\right)\cdot\left(x^2+4x+5\right)\)

\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+5\right)\)

\(=x\left(x^2+2xy+y^2-9\right)\)

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

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