Question

Cho f(x) =ax^2+bx+c. Tính f(x+3) -3f(x+2) +3f(x+1) -f(x)

Answer 1

Lời giải:

Ta có: \(f(x)=ax^2+bx+c\)

\(\Rightarrow \left\{\begin{matrix} f(x+3)=a(x+3)^2+b(x+3)+c\\ f(x+2)=a(x+2)^2+b(x+2)+c\\ f(x+1)=a(x+1)^2+b(x+1)+c\\ f(x)=ax^2+bx+c\end{matrix}\right.\)

\(\Rightarrow f(x+3)-3f(x+2)+3f(x+1)-f(x)\)

\(=[f(x+3)-f(x)]-3[f(x+2)-f(x+1)]\)

Có:

\(f(x+3)-f(x)=a(x+3)^2+b(x+3)+c-[ax^2+bx+c]\)

\(=a[(x+3)^2-x^2]+b(x+3-x)\)

\(=3a(2x+3)+3b(1)\)

Và: \(f(x+2)-f(x+1)=a[(x+2)^2-(x+1)^2]+b[(x+2)-(x+1)]\)

\(=a(2x+3)+b\)

\(\Rightarrow 3[f(x+2)-f(x+1)]=3a(2x+3)+3b(2)\)

Từ (1)(2) suy ra:

\(f(x+3)-3f(x+2)+3f(x+1)-f(x)=3a(2x+3)+3b-[3a(2x+3)+3b]=0\)