Ta có \(f\left(7\right)=15\Rightarrow f\left(7\right)-15=0\Rightarrow f\left(x\right)-15=P\left(x\right).\left(x-7\right)\)

\(\Rightarrow f\left(15\right)-15=P\left(x\right).8\Rightarrow-15=P\left(x\right).8\Rightarrow P\left(x\right)=\dfrac{-3}{4}\). (vô lí vì P(x) có các hệ số đều nguyên).

Vậy...

- Nếu \(a_i=0\) ; \(\forall i\in\left(0;n-1\right)\Rightarrow a_nx^n=0\Rightarrow\alpha=0< 1\) thỏa mãn

- Nếu tồn tại \(a_i\ne0\), đặt \(max\left|\dfrac{a_i}{a_n}\right|=A>0\)

Do \(\alpha\) là nghiệm nên:

\(a_n\alpha^n+a_{n-1}\alpha^{n-1}+...+a_1\alpha+a_0=0\)

\(\Leftrightarrow\dfrac{a_0}{a_n}+\dfrac{a_1}{a_n}\alpha+...+\dfrac{a_{n-1}}{a_n}\alpha^{n-1}=-\alpha^n\)

\(\Leftrightarrow\left|\alpha^n\right|=\left|\dfrac{a_0}{a_n}+\dfrac{a_1}{a_n}\alpha+...+\dfrac{a_{n-1}}{a_n}\alpha^{n-1}\right|\)

\(\Rightarrow\left|\alpha^n\right|\le\left|\dfrac{a_0}{a_n}\right|+\left|\dfrac{a_1}{a_n}\right|.\left|\alpha\right|+...+\left|\dfrac{a_{n-1}}{a_n}\right|.\left|\alpha^{n-1}\right|\le A+A.\left|\alpha\right|+...+A.\left|\alpha^{n-1}\right|\)

\(\Rightarrow\left|\alpha^n\right|\le A\left(1+\left|\alpha\right|+\left|\alpha^2\right|+...+\left|\alpha^{n-1}\right|\right)\)

\(\Rightarrow\left|\alpha^n\right|\le A.\dfrac{\left|\alpha^n\right|-1}{\left|\alpha\right|-1}\)

TH1: Nếu \(\left|\alpha\right|\le1\) hiển nhiên ta có \(\left|\alpha\right|< 1+A\) (đpcm)

TH2: Nếu \(\left|\alpha\right|>1\)

\(\Rightarrow\left|\alpha^n\right|\le\dfrac{A.\left|\alpha^n\right|}{\left|\alpha\right|-1}-\dfrac{A}{\left|\alpha\right|-1}< \dfrac{A.\left|\alpha^n\right|}{\left|\alpha\right|-1}\)

\(\Leftrightarrow\left|\alpha\right|-1< A\Rightarrow\left|\alpha\right|< 1+A\) (đpcm)

a. Ta có: f(x) + h(x) = g(x)

Suy ra: h(x) = g(x) – f(x) = (x4 – x3 + x2 + 5) – (x4 – 3x2 + x – 1)

= x4 – x3 + x2 + 5 – x4 + 3x2 – x + 1

= -x3 + 4x2 – x + 6

b. Ta có: f(x) – h(x) = g(x)

Suy ra: h(x) = f(x) – g(x) = (x4 – 3x2 + x – 1) – (x4 – x3 + x2 + 5)

= x4 – 3x2 + x – 1 – x4 + x3 – x2 – 5

= x3 – 4x2 + x – 6