Cho đa thức \(f(x)={{a}_{4}}{{x}^{4}}+{{a}_{3}}{{x...

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Theo đề bài ta có:

\(\begin{align}& f(1)={{a}_{4}}{{.1}^{4}}+{{a}_{3}}{{.1}^{3}}+{{a}_{2}}{{.1}^{2}}+{{a}_{1}}.1+{{a}_{0}}={{a}_{4}}+{{a}_{3}}+{{a}_{2}}+{{a}_{1}}+{{a}_{0}} \\  & f(-1)={{a}_{4}}.{{(-1)}^{4}}+{{a}_{3}}.{{(-1)}^{3}}+{{a}_{2}}.{{(-1)}^{2}}+{{a}_{1}}.1+{{a}_{0}}={{a}_{4}}-{{a}_{3}}+{{a}_{2}}-{{a}_{1}}+{{a}_{0}} \\ \end{align}\)

Vì \(f(1)=f(-1)\) nên ta có:

\(\begin{align}& {{a}_{4}}+{{a}_{3}}+{{a}_{2}}+{{a}_{1}}+{{a}_{0}}={{a}_{4}}-{{a}_{3}}+{{a}_{2}}-{{a}_{1}}+{{a}_{0}} \\ & \Rightarrow {{a}_{3}}+{{a}_{1}}=-{{a}_{3}}-{{a}_{1}}\,\Leftrightarrow 2{{a}_{3}}+2{{a}_{1}}=0\,\,\Leftrightarrow \,{{a}_{3}}+{{a}_{1}}=0\,\Leftrightarrow {{a}_{3}}=-{{a}_{1}}\,\,\,\,\,(1) \\ \end{align}\)

\(\begin{align}& f(1)={{a}_{4}}{{.2}^{4}}+{{a}_{3}}{{.2}^{3}}+{{a}_{2}}{{.2}^{2}}+{{a}_{1}}.2+{{a}_{0}}=16{{a}_{4}}+8{{a}_{3}}+4{{a}_{2}}+2{{a}_{1}}+{{a}_{0}} \\ & f(-2)={{a}_{4}}.{{(-2)}^{4}}+{{a}_{3}}.{{(-2)}^{3}}+{{a}_{2}}.{{(-2)}^{2}}+{{a}_{1}}.2+{{a}_{0}}=14{{a}_{4}}-8{{a}_{3}}+4{{a}_{2}}-2{{a}_{1}}+{{a}_{0}} \\ \end{align}\)

Vì \(f(2)=f(-2)\) nên ta có:

\(\begin{align}  & 16{{a}_{4}}+8{{a}_{3}}+4{{a}_{2}}+2{{a}_{1}}+{{a}_{0}}=16{{a}_{4}}-8{{a}_{3}}+4{{a}_{2}}-2{{a}_{1}}+{{a}_{0}} \\  & \Rightarrow 8{{a}_{3}}+2{{a}_{1}}=-8{{a}_{3}}-2{{a}_{1}}\,\Leftrightarrow 16{{a}_{3}}+4{{a}_{1}}=0\,\,\Leftrightarrow 4{{a}_{3}}+{{a}_{1}}=0\,\,\,\,\,\,(2) \\ \end{align}\)

Thế (1) vào (2) ta được:  \(4{{a}_{3}}-{{a}_{3}}=0\Leftrightarrow {{a}_{3}}=0\Rightarrow {{a}_{3}}={{a}_{1}}=0.\)

Vậy đa thức \(f(x)={{a}_{4}}{{x}^{4}}+{{a}_{2}}{{x}^{2}}+{{a}_{0}}\).

Vì \({{x}^{4}}={{(-x)}^{4}}\,;\,\,{{x}^{2}}={{(-x)}^{2}}\) với mọi x, do đó \({{a}_{4}}{{x}^{4}}+{{a}_{2}}{{x}^{2}}+{{a}_{0}}={{a}_{4}}{{(-x)}^{4}}+{{a}_{2}}{{(-x)}^{2}}+{{a}_{0}}\)

Suy ra \(f(x)=f(-x)\) với mọi x.