Question

Phân tích các đa thức sau thành nhân tử:

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

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

c) \(3x^4y^2+3x^3y^2+3xy^2+3y^2\)

Answer 1

\(c,3x^4y^2+3x^3y^2+3xy^2+3y^2\)

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

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

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

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

Answer 2

a, \(\left(xy+1\right)^2-\left(x+y\right)^2\)

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

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

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

b,\(3x^4y^2+3x^3y^2+3xy^2+3y^2\)

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

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

Answer 3

a. (xy+1)² - (x+y)²

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

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

Answer 4

a, Theo đề ra ta có:

= [ (xy)²- 2xy + 1 ]- ( x² + 2xy + y² )

= x²y² + 2xy + 1 - x² - 2xy - y²

= x²y² + 1 - x² - y²

= ( y² - 1 ). ( x²y² - x² )

= ( y² - 1 ). x²( y² - 1 )

=>. ( y² - 1 ). ( x² - 1 )

Answer 5

b, theo đề ra ta có:

= ( x³ + 3x²y + 3xy² + y³ ) - ( x³ - 3x²y + 3xy² - y³ )

= x³ + 3x²y + 3xy² + y³ - x³ + 3x²y - 3xy² + y³

= 6x⁴y²