Đáp án:
$B=2$
$D=3$
Giải thích các bước giải:
Dạng $2$
a.Ta có:
$B=(x^2-2)(x^2+x-1)-x(x^3+x^2-3x-2)$
$\to B=x^4+x^3-3x^2-2x+2-(x^4+x^3-3x^2-2x)$
$\to B=x^4+x^3-3x^2-2x+2-x^4-x^3+3x^2+2x$
$\to B=x^4-x^4+x^3-x^3-3x^2+3x^2-2x+2x+2$
$\to B=2$
b.Ta có:
$D=x(2x+1)-x^2(x+2)+x^3-x+3$
$\to D=2x^2+x-(x^3+2x^2)+x^3-x+3$
$\to D=2x^2+x-x^3-2x^2+x^3-x+3$
$\to D=-x^3+x^3+2x^2-2x^2+x-x+3$
$\to D=3$
Dạng $3$
a.Ta có:
$VT=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$
$\to VT=x(x^4+x^3y+x^2y^2+xy^3+y^4)-y(x^4+x^3y+x^2y^2+xy^3+y^4)$
$\to VT=x^5+x^4y+x^3y^2+x^2y^3+xy^4-(yx^4+y^2x^3+y^3x^2+y^4x+y^5)$
$\to VT=x^5+x^4y+x^3y^2+x^2y^3+xy^4-yx^4-y^2x^3-y^3x^2-y^4x-y^5$
$\to VT=x^5+x^4y-x^4y+x^3y^2-x^3y^2+x^2y^3-x^2y^3+xy^4-xy^4-y^5$
$\to VT=x^5-y^5$
$\to VT=VP$
$\to đpcm$
b.Ta có:
$VT=(a+b)(a^3-a^2b+ab^2-b^3)$
$\to VT=a(a^3-a^2b+ab^2-b^3)+b(a^3-a^2b+ab^2-b^3)$
$\to VT=a^4-a^3b+a^2b^2-ab^3+ba^3-a^2b^2+ab^3-b^4$
$\to VT=a^4-a^3b+a^3b+a^2b^2-a^2b^2-ab^3+ab^3-b^4$
$\to VT=a^4-b^4$
$\to VT=VP$
$\to đpmc$
