Bài 2
a) Ta có
$A = x^2 +9y^2 - 6xy$
$= x^2 +(3y)^2 + 2.x.3y$
$= (x-3y)^2$
Thay $x = 19, y = 3$ vào ta có
$A = (19-3.3)^2 = 10^2 = 100$
b) Ta có
$B = x^3 - 6x^2y + 12xy^2 -8y^3$
$= x^3 - 3.x^2.(2y) + 3.x.(2y)^2 - (2y)^3$
$= (x-2y)^3$
Thay $x = 12, y = -4$ vào ta có
$B = (12 -2(-4))^3 = 20^3 = 8000$
Bài 3
a) $3(x-y)^2 - 2(x+y)^2 - (x-y)(x+y) = 3(x-y)^2 -3(x-y)(x+y) + 2(x-y)(x+y) - 2(x+y)^2$
$= 3(x-y)[(x-y)-(x+y)] + 2(x+y)[(x-y)-(x+y)]$
$= 3(x-y)(-2y) + 2(x+y)(-2y)$
$= (-2y)[3(x-y)+2(x+y)]$
$=-2y(5x-y)$
c) $(x-4)^2-2(x-4)(x+5) + (x+5)^2 = (x-4)^2 - (x-4)(x+5) - (x-4)(x+5) + (x+5)^2$
$= (x-4)[(x-4)-(x+5)] - (x+5)[(x-4)-(x+5)]$
$= (x-4)(-9) - (x+5)(-9)$
$= (-9)(x-4-x-5)$
$= -9(-9) = 81$
