Giải thích các bước giải:
Bài 1:
a.Ta có:
$A(x)=9-x^5+4x-2x^3+x^2-7x^4$
$\to A(x)=-x^5-7x^4-2x^3+x^2+4x+9$
$B(x)=x^5-9+2x^2+7x^4+2x^3-3x$
$\to B(x)=x^5+7x^4+2x^3+2x^2-3x-9$
b.Ta có:
$H(x)=A(x)+B(x)$
$\to H(x)=( -x^5-7x^4-2x^3+x^2+4x+9)+(x^5+7x^4+2x^3+2x^2-3x-9)$
$\to H(x)=-x^5-7x^4-2x^3+x^2+4x+9+x^5+7x^4+2x^3+2x^2-3x-9$
$\to H(x)=-x^5+x^5-7x^4+7x^4-2x^3+2x^3+x^2+2x^2+4x-3x+9-9$
$\to H(x)=3x^2+x$
c.Ta có:
$G(x)=B(x)-A(x)$
$\to G(x)=(x^5+7x^4+2x^3+2x^2-3x-9)-( -x^5-7x^4-2x^3+x^2+4x+9)$
$\to G(x)=x^5+7x^4+2x^3+2x^2-3x-9+x^5+7x^4+2x^3-x^2-4x-9$
$\to G(x)=x^5+x^5+7x^4+7x^4+2x^3+2x^3+2x^2-x^2-3x-4x-9-9$
$\to G(x)=2x^5+14x^4+4x^3+x^2-7x-18$
Bài 2:
a.Ta có:
$C(x)=-3x^2+x-1+x^4-x^3-x^2+3x^4$
$\to C(x)=x^4+3x^4-x^3-3x^2-x^2+x-1$
$\to C(x)=4x^4-x^3-4x^2+x-1$
$D(x)=x^4+x^2-x^3+x-5+5x^3-x^2$
$\to D(x)=x^4-x^3+5x^3+x^2-x^2+x-5$
$\to D(x)=x^4+4x^3+x-5$
b.Ta có:
$C(x)-D(x)$
$=(4x^4-x^3-4x^2+x-1)-(x^4+4x^3+x-5)$
$=4x^4-x^3-4x^2+x-1-x^4-4x^3-x+5$
$=4x^4-x^4-x^3-4x^3-4x^2+x-x-1+5$
$=3x^4-5x^3-4x^2+4$
Ta có:
$C(x)+D(x)$
$=(4x^4-x^3-4x^2+x-1)+(x^4+4x^3+x-5)$
$=4x^4-x^3-4x^2+x-1+x^4+4x^3+x-5$
$=4x^4+x^4+3x^3-4x^2+x+x-1-5$
$=5x^4+3x^3-4x^2+2x-6$
