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

a) Ta có:

f(x)+g(x)=a0xn+a1xn−1+a2xn−2+…+an−1x+an+b0xn+b1xn−1+b2xn−2 +…+bn−1x+bn

=(a0+b0)xn+(a1+b1)xn−1+(a2+b2)xn−2+…+(an−1+bn−1)x+(an+bn)

b) Ta có

f(x)+g(x)=a0xn+a1xn−1+a2xn−2+…+an−1x+an-b0xn+b1xn−1+b2xn−2 +…+bn−1x+bn

=a0xn+a1xn−1+a2xn−2+…+an−1x+an−b0xn−b1xn−1−b2xn−2−…−bn−1x+bn

=(a0−b0)xn+(a1−b1)xn−1+(a2−b2)xn−2+…+(an−1−bn−1)x+(an−bn)