\(c,3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=3x^3y^2\left(x+1\right)+3y^2\left(x+1\right)\)
\(=3y^2\left(x+1\right)\left(x^3+1\right)\)
\(=3y^2\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
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,\(3x^4y^2+3x^3y^2+3xy^2+3y^2\)
= \(\left(3x^4y^2+3x^3y^2\right)+\left(3xy^2+3y^2\right)\)
= \(3x^3y^2\left(x+1\right)+3y^2\left(x+1\right)=3y^2\left(x+1\right)\left(x^3+1\right)\)
a. (xy+1)2 - (x+y)2
= (xy+1+x+y) (xy+1-x-y)
= (x+1)(y+1)(x-1)(y-1)
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 )
b, theo đề ra ta có:
= ( x3 + 3x2y + 3xy2 + y3 ) - ( x3 - 3x2y + 3xy2 - y3 )
= x3 + 3x2y + 3xy2 + y3 - x3 + 3x2y - 3xy2 + y3
= 6x4y2
