A(x)=3x^6-x^5-3x^3+x^2+x-1;B(x)=-x^5-3x^3+2x^2+6x^5+x-15 tim da thuc C(x)+B(x)=A(x) 27/10/2021 Bởi Margaret A(x)=3x^6-x^5-3x^3+x^2+x-1;B(x)=-x^5-3x^3+2x^2+6x^5+x-15 tim da thuc C(x)+B(x)=A(x)
C(x)+B(x)=A(x) C(x)+($-x^{5}$ -$3x^{3}$ +$2x^{2}$ +x-15)=$3x^{6}$ -$x^{5}$ -$3x^{3}$ +$x^{2}$ +x-1 C(x)=$3x^{6}$ -$x^{5}$ -$3x^{3}$ +$x^{2}$ +x-1-($-x^{5}$ -$3x^{3}$ +$2x^{2}$ +x-15) C(x)=$3x^{6}$ -$x^{5}$ -$3x^{3}$ +$x^{2}$ +x-1$+x^{5}$ +$3x^{3}$ -$2x^{2}$ -x+15 C(x)=$3x^{6}$+(-$x^{5}$+$x^{5}$ )+(-$3x^{3}$+$3x^{3}$)+($x^{2}$-$2x^{2}$)+(x-x)+(-1+15) C(x)=$3x^{6}$-$x^{2}$ +14 Vậy C(x)=$3x^{6}$-$x^{2}$ +14 Bình luận
$A(x)=3x^6-x^5-3x^3+x^2+x-1$ $B(x)=-x^5-3x^3+2x^2+x-15$ $C(x)+B(x)=A(x)$ $⇒C(x)=A(x)-B(x)$ $⇒C(x)=(3x^6-x^5-3x^3+x^2+x-1)-(-x^5-3x^3+2x^2+x-15)$ $⇒C(x)=3x^6-x^5-3x^3+x^2+x-1+x^5+3x^3-2x^2-x+15$ $⇒C(x)=3x^6+(-x^5+x^5)+(-3x^3+3x^3)+(x^2-2x^2)+(x-x)+(-1+15)$ $⇒C(x)=3x^6-x^2+14$ Vậy $C(x)=3x^6-x^2+14$. Bình luận
C(x)+B(x)=A(x)
C(x)+($-x^{5}$ -$3x^{3}$ +$2x^{2}$ +x-15)=$3x^{6}$ -$x^{5}$ -$3x^{3}$ +$x^{2}$ +x-1
C(x)=$3x^{6}$ -$x^{5}$ -$3x^{3}$ +$x^{2}$ +x-1-($-x^{5}$ -$3x^{3}$ +$2x^{2}$ +x-15)
C(x)=$3x^{6}$ -$x^{5}$ -$3x^{3}$ +$x^{2}$ +x-1$+x^{5}$ +$3x^{3}$ -$2x^{2}$ -x+15
C(x)=$3x^{6}$+(-$x^{5}$+$x^{5}$ )+(-$3x^{3}$+$3x^{3}$)+($x^{2}$-$2x^{2}$)+(x-x)+(-1+15)
C(x)=$3x^{6}$-$x^{2}$ +14
Vậy C(x)=$3x^{6}$-$x^{2}$ +14
$A(x)=3x^6-x^5-3x^3+x^2+x-1$
$B(x)=-x^5-3x^3+2x^2+x-15$
$C(x)+B(x)=A(x)$
$⇒C(x)=A(x)-B(x)$
$⇒C(x)=(3x^6-x^5-3x^3+x^2+x-1)-(-x^5-3x^3+2x^2+x-15)$
$⇒C(x)=3x^6-x^5-3x^3+x^2+x-1+x^5+3x^3-2x^2-x+15$
$⇒C(x)=3x^6+(-x^5+x^5)+(-3x^3+3x^3)+(x^2-2x^2)+(x-x)+(-1+15)$
$⇒C(x)=3x^6-x^2+14$
Vậy $C(x)=3x^6-x^2+14$.