Question

Phân tích đa thức thành nhân tử

x²-2xy+y²+3x-3y-10

Answer 1

\(x^2-2xy+y^2+3x-3y-10=\left(x^2-2xy+y^2\right)+3\left(x-y\right)-10=\left(x-y\right)^2+3\left(x-y\right)-10=\left(x-y\right)^2-2\left(x-y\right)+5\left(x-y\right)-10=\left(x-y\right)\left(x-y-2\right)+5\left(x-y-2\right)=\left(x-y+5\right)\left(x-y-2\right)\)

Answer 2

= ( x - y)^2  - 3 ( x - y) . -10

= ( x  - y)^2 - 2.(x-y) . 3/2 +9/4 - 49/4 

= ( x - y - 3/2) ^2 - (7/2)^2 

= ( x- y - 3/2 - 7/2 )( x - y -3/2 + 7/2 )

=( x - y - 5 )( x - y + 2)