Đáp án:
Giải thích các bước giải:
bài 1:
$a)6x^2.(3x^2-4x+5)$
$=18x^4-24x^3+30x^2$
$b)(x-2y).(3xy+6y^2+x)$
$=3x^2y+6xy^2+x^2-6xy^2-12y^3-2xy$
$=-12y^3+3x^2y+x^2-2xy$
$c)(18x^4y^3-24x^3y^4+12x^3y^3):(-6x^2y^3)$
$=-3x^2+4xy-2x$
bài 2:
$a)3x^2-3xy-5x+5y$
$=3x.(x-y)-5.(x-y)$
$=(x-y).(3x-5)$
$b)x^2+4x-45$
$=x^2+4x+4-49$
$=(x+2)^2-7^2$
$=(x+2-7).(x+2+7)$
$=(x-5).(x+9)$
$c)3y^3+6xy^2+3x^2y$
$=3y.(y^2+2xy+x^2)$
$=3y.(x+y)^2$
$d)x^3-3x^2-4x+12$
$=(x^3-3x^2)-(4x-12)$
$=x^2.(x-3)-4.(x-3)$
$=(x-3).(x^2-4)$
$=(x-3).(x-2).(x+2)$
$e)x^3+3x^2-3x-1$
$=(x^3-1)+(3x^2-3x)$
$=(x-1).(x^2+x+1)+3x.(x-1)$
$=(x-1).(x^2+x+1+3x)$
$=(x-1).(x^2+4x+1)$
$f)x^2-3x+xy-3y$
$=(x^2+xy)-(3x+3y)$
$=x.(x+y)-3.(x+y)$
$=(x+y).(x-3)$
$g)x^2-2xy+y^2-4$
$=(x-y)^2-2^2$
$=(x-y-2).(x-y+2)$
$h)x^2-2xy+y^2-z^2$
$=(x^2-2xy+y^2)-z^2$
$=(x-y)^2-z^2$
$=(x-y-z).(x-y+z)$
$i)3x^2+6xy+3y^2-3z^2$
$=3.[ x^2+2xy+y^2-z^2]$
$=3.[ (x^2+2xy+y^2)-z^2]$
$=3.[ (x+y)^2-z^2]$
$=3.(x+y-z).(x+y+z)$