Đáp án:
a.$(x+y+1)(2x-y+2)$
b.$(3x-y+2)(x-3y+1)$
Giải thích các bước giải:
a.Ta có:
$2x^2-y^2+xy+4x+y+2$
$=(2x^2+4x+2)-y^2+(xy+y)$
$=2(x^2+2x+1)-y^2+y(x+1)$
$=2(x+1)^2-y^2+y(x+1)$
$=(2(x+1)^2+2y(x+1))-(y^2+y(x+1))$
$=2(x+1)(x+1+y)-y(y+x+1)$
$=2(x+1)(x+y+1)-y(x+y+1)$
$=(x+y+1)(2(x+1)-y)$
$=(x+y+1)(2x-y+2)$
b.Ta có:
$3x^2+3y^2-10xy+5x-7y+2$
$=(3x^2-9xy+3x)+(3y^2-xy-y)+(2x-6y+2)$
$=3x(x-3y+1)+y(3y-x-1)+2(x-3y+1)$
$=3x(x-3y+1)-y(x-3y+1)+2(x-3y+1)$
$=(3x-y+2)(x-3y+1)$
