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)

Question

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)

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Margaret 1 tháng 2021-10-27T07:44:14+00:00 2 Answers 7 views 0

Answers ( )

    0
    2021-10-27T07:45:44+00:00

    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

     

    0
    2021-10-27T07:46:09+00:00

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

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

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

    $⇒C(x)=A(x)-B(x)$

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

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

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

    $⇒C(x)=3x^6-x^2+14$

    Vậy $C(x)=3x^6-x^2+14$.

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