Cho hai đa thức sau: f(x) = a0 x^n + a1 x^n−1 + a2 x^n−2 + … + an−1

* Hướng dẫn giải

a) Ta có:

$f\left(x\right)+g\left(x\right)=a_0{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}+\dots +a_{n-1}x+a_n+b_0{x}^n+b_1{x}^{n-1}+b_2{x}^{n-2} +\dots +b_{n-1}x+b_n$

$=(a_0+b_0)x^n+(a_1+b_1)x^{n-1}+(a_2+b_2)x^{n-2}+\dots +(a_{n-1}+b_{n-1})x+(a_n+b_n)$

b) Ta có

$f\left(x\right)+g\left(x\right)=a_0{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}+\dots +a_{n-1}x+a_n-\left(b_0{x}^n+b_1{x}^{n-1}+b_2{x}^{n-2} +\dots +b_{n-1}x+b_n\right)$

$=a_0{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}+\dots +a_{n-1}x+a_n-b_0{x}^n-b_1{x}^{n-1}-b_2{x}^{n-2}-\dots -b_{n-1}x+b_n$

$=(a_0-b_0)x^n+(a_1-b_1)x^{n-1}+(a_2-b_2)x^{n-2}+\dots +(a_{n-1}-b_{n-1})x+(a_n-b_n)$