Đáp án + Giải thích các bước giải:
$a)(3x+1)^2-4(x-2)^2=[3x+1-2(x-2)][3x+1+2(x-2)]$
$=(3x+1-2x+4)(3x+1+2x-4)=(x+5)(5x-3)$
$b)9(2x+3)^2-4(x+1)^2=[3(2x+3)]^2-[2(x+1)]^2$
$=[3(2x+3)-2(x+1)][3(2x+3)+2(x+1)]$
$=(6x+9-2x-2)(6x+9+2x+2)=(4x+7)(8x+11)$
$c)4b^2c^2-(b^2+c^2-a^2)^2=(2bc)^2-(b^2+c^2-a^2)^2$
$=(2bc-b^2-c^2+a^2)(2bc+b^2+c^2-a^2)$
$=-[(b^2-2bc+c^2)-a^2][(b^2+2bc+c^2)-a^2]$
$=-[(b-c)^2-a^2][(b+c)^2-a^2]$
$=-(b-c-a)(b-c+a)(b+c-a)(b+c+a)$
$d)(xy+1)^2-(x+y)^2=(xy+1-x-y)(xy+1+x+y)$
$e)(x+y)^3-(x-y)^3=(x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]$
$=2y(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2)$
$=2y(3x^2+y^2)=6x^2y+2y^3$
Mk xin hya nhất ạ
