Cho đa thức : `P(x) x^5 – 3x^2+x^4 – 1/2x-x^5+5x^4+x^2-1` `Q(x)=x-x^9+x^2 – 5-x^3+x^6-x+3x^9+2x^6-x^3+7` Tính : `P(x) + Q(x)`

1 bình luận về “Cho đa thức : `P(x) x^5 – 3x^2+x^4 – 1/2x-x^5+5x^4+x^2-1` `Q(x)=x-x^9+x^2 – 5-x^3+x^6-x+3x^9+2x^6-x^3+7` Tính : `P(x) + Q(x)`”

1. Ta có:

   @ P(x)= x^5- 3x^2+ x^4- 1/2x- x^5+ 5x^4+ x^2- 1

   = (x^5- x^5)+ (x^4+ 5x^4)- (3x^2- x^2) – 1/2x- 1

   = 6x^4- 2x^2- 1/2x- 1

   @ Q(x)= x- x^9+ x^2- 5- x^3+ x^6- x+ 3x^9+ 2x^6- x^3+7

   = (3x^9- x^9)+ (x^6+ 2x^6)- (x^3+ x^3)+ x^2+ (x-x)+(7-5)

   = 2x^9+ 3x^6- 2x^3+ x^2+2

   _

   ⇒ P(x)+ Q(x)= (6x^4- 2x^2- 1/2x- 1)+(2x^9+ 3x^6- 2x^3+ x^2+2)

   = 6x^4- 2x^2- 1/2x- 1+  2x^9+ 3x^6- 2x^3+ x^2+2 

   = 2x^9+ 3x^6+ 6x^4- 2x^3- x^2- 1/2x+ 1

   Vậy, P(x)+ Q(x)= 2x^9+ 3x^6+ 6x^4- 2x^3- x^2- 1/2x+ 1

   ~ $kiddd$ ~

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