Đặt [tex]F(f(x))[/tex] là tổng hệ số của đa thức f(x), [tex]f(x)=a_{1}x^n+a_2x^{n-1}+...+a_{n+1},g(x)=b_{1}x^{n+i}+b_2x^{n+i-1}+...+b_{n+i+1}[/tex]
Ta cần chứng minh:
+ [tex]F(f(x)+g(x))=F(f(x))+F(g(x))[/tex]
Thật vậy, [tex]f(x)+g(x)=b_1x^{n+i}+b_2x^{n+i-1}+...+b_ix_{n+1}+(a_1+b_{i+1})x^n+...+(a_n+b_{n+i+1})[/tex]
Khi đó [tex]F(f(x)+g(x))=b_1+b_2+...+b_i+(a_1+b_{i+1})+...+(a_n+b_{n+i+1})=F(f(x))+F(g(x))[/tex]
+ [tex]F(f(x)).F(g(x))=F(f(x).g(x))[/tex]
Thật vậy, [tex]F(f(x).g(x))=F((a_1x^n+a_2x^{n-1}+...+a_{n+1})(b_1x^{n+i}+...+b_{n+i+1}))=F(a_1x^n(b_1x^{n+i}+...+b_{n+i+1})+a_2x^{n-1}(b_1x^{n+i}+...+b_{n+i+1})+...+a_{n+1}(b_1x^{n+i}+...+b_{n+i+1}))=F(a_1x^n(b_1x^{n+i}+...+b_{n+i+1}))+F(a_2x^{n-1}(b_1x^{n+i}+...+b_{n+i+1}))+...+F(a_{n+1}(b_1x^{n+i}+...+b_{n+i+1}))=a_1(b_1+b_2+...+b_{n+i})+a_2(b_1+b_2+...+b_{n+i+1})=(a_1+a_2+...+a_n)(b_1+b_2+...+b_{n+i+1})=F(f(x)).F(g(x))[/tex]
Ta có: [tex]F(f(g(x)))=F(a_1(b_{1}x^{n+i}+b_2x^{n+i-1}+...+b_{n+i+1})^n+...+a_{n+1})=F(a_1(b_1+b_2+...+b_{n+i+1}))+...+F(a_{n+1})=a_1(b_1+b_2+...+b_{n+i+1})^n+...+a_{n+1}=F(f(g(1)))[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
