Question

Phân tích đa thức thành nhân tử 

x^2 +y^2 +2xy +x +y -6

Answer 1

Ta có: \(x^2+y^2+2xy+x+y-6\)

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

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

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

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

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

Answer 2

ta có : x² + y² + 2xy + x + y - 6

= ( x + y ) ² + x + y - 6

= ( x + y ) ² + x + y - 9 + 3 

=[ ( x + y )² - 3² ] + ( x + y + 3 )

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

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