Bài toán: Cho đa thức bậc $n$ có $n$ nghiệm phân biệt là$x_1,x_2,...x_n$.Chứng minh rằng
\[\frac{{P"({x_1})}}{{P'({x_1})}} + \frac{{P"({x_2})}}{{P'({x_2})}} + ... + \frac{{P"({x_n})}}{{P'({x_n})}} = 0\]
Lời giải:
Xét $P\left( x \right) = \left( {x - {x_1}} \right)\left( {x - {x_2}} \right)...\left( {x - {x_n}} \right)$, giả sử ${x_1} < {x_2} < ... < {x_n}$
Ta có: $$P'\left( x \right) = P\left( x \right)\left( {\dfrac{1}{{x - {x_1}}} + \dfrac{1}{{x - {x_2}}} + ... + \dfrac{1}{{x - {x_n}}}} \right) = P\left( x \right)\sum\limits_{i = 1}^n {\dfrac{1}{{x - {x_i}}}} \,\,\,\,\,\left( 1 \right)$$
Do $P\left( {{x_i}} \right) = 0,i = \overline {1,n} $ nên theo định lí Rolle tồn tại ${c_1},{c_2},...,{c_{n - 1}};\,\,{x_1} < {c_1} < {x_2} < {c_2} < ... < {c_{n - 1}} < {x_n}$ sao cho $P'\left( {{c_i}} \right) = 0,i = \overline {1,n - 1} \,\,\,\,\,\,\,\,\,\,\left( 2 \right)$
Lại có: $$P''\left( x \right) = P'\left( x \right)\left( {\dfrac{1}{{x - {c_1}}} + \dfrac{1}{{x - {c_2}}} + ... + \dfrac{1}{{x - {c_{n - 1}}}}} \right) = P'\left( x \right)\sum\limits_{i = 1}^{n - 1} {\dfrac{1}{{x - {c_i}}}} \,\,\,\,\,\left( 3 \right)$$
Từ (1) và (2) suy ra:$$\left\{ \begin{array}{l}
P'\left( {{c_1}} \right) = P\left( {{c_1}} \right)\left( {\dfrac{1}{{{c_1} - {x_1}}} + \dfrac{1}{{{c_1} - {x_2}}} + ... + \dfrac{1}{{{c_1} - {x_n}}}} \right) = P\left( {{c_1}} \right)\sum\limits_{i = 1}^n {\dfrac{1}{{{c_1} - {x_i}}} = 0} \\
P'\left( {{c_2}} \right) = P\left( {{c_2}} \right)\left( {\dfrac{1}{{{c_2} - {x_1}}} + \dfrac{1}{{{c_2} - {x_2}}} + ... + \dfrac{1}{{{c_2} - {x_n}}}} \right) = P\left( {{c_2}} \right)\sum\limits_{i = 1}^n {\dfrac{1}{{{c_2} - {x_i}}} = 0} \\
...............\\
P'\left( {{c_{n - 1}}} \right) = P\left( {{c_{n - 1}}} \right)\left( {\dfrac{1}{{{c_{n - 1}} - {x_1}}} + \dfrac{1}{{{c_{n - 1}} - {x_2}}} + ... + \dfrac{1}{{{c_{n - 1}} - {x_n}}}} \right) = P\left( {{c_{n - 1}}} \right)\sum\limits_{i = 1}^n {\dfrac{1}{{{c_{n - 1}} - {x_i}}} = 0}
\end{array} \right.$$
Do $P\left( {{c_i}} \right) \ne 0,i = \overline {1,n - 1} $ nên ta có:$$\left\{ \begin{array}{l}
\dfrac{1}{{{c_1} - {x_1}}} + \dfrac{1}{{{c_1} - {x_2}}} + ... + \dfrac{1}{{{c_1} - {x_n}}} = \sum\limits_{i = 1}^n {\dfrac{1}{{{c_1} - {x_i}}} = 0} \\
\dfrac{1}{{{c_2} - {x_1}}} + \dfrac{1}{{{c_2} - {x_2}}} + ... + \dfrac{1}{{{c_2} - {x_n}}} = \sum\limits_{i = 1}^n {\dfrac{1}{{{c_2} - {x_i}}} = 0} \\
..............\\
\dfrac{1}{{{c_{n - 1}} - {x_1}}} + \dfrac{1}{{{c_{n - 1}} - {x_2}}} + ... + \dfrac{1}{{{c_{n - 1}} - {x_n}}} = \sum\limits_{i = 1}^n {\dfrac{1}{{{c_{n - 1}} - {x_i}}} = 0}
\end{array} \right.$$
Suy ra: $$\sum\limits_{i = 1}^n {\dfrac{{P''\left( {{x_i}} \right)}}{{P'\left( {{x_i}} \right)}} = } \sum\limits_{i = 1}^n {\dfrac{1}{{{c_1} - {x_i}}} + \sum\limits_{i = 1}^n {\dfrac{1}{{{c_2} - {x_i}}} + \sum\limits_{i = 1}^n {\dfrac{1}{{{c_{n - 1}} - {x_i}}} = 0} } } $$
Bài toán được chứng minh.