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)

0 bình luận về “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)”

1. 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

    

   Trả lời

2. $A(x)=3x^6-x^5-3x^3+x^2+x-1$

   $B(x)=-x^5-3x^3+2x^2+x-15$

   $C(x)+B(x)=A(x)$

   $⇒C(x)=A(x)-B(x)$

   $⇒C(x)=(3x^6-x^5-3x^3+x^2+x-1)-(-x^5-3x^3+2x^2+x-15)$

   $⇒C(x)=3x^6-x^5-3x^3+x^2+x-1+x^5+3x^3-2x^2-x+15$

   $⇒C(x)=3x^6+(-x^5+x^5)+(-3x^3+3x^3)+(x^2-2x^2)+(x-x)+(-1+15)$

   $⇒C(x)=3x^6-x^2+14$

   Vậy $C(x)=3x^6-x^2+14$.

   Trả lời