Question

f=x^4- 3x^2+x+7 apartine R[X]
Calculati $\frac{1}{2-x1}$ + $\frac{1}{2-x2}$ + $\frac{1}{2-x3}$ + $\frac{1}{2-x4}$


Nu mai am puncte
Ajutorrrr bacccc

Answer 1

$f(x) = x^4-3x^2+x+7\\ \\f'(x) = 4x^3-6x+1\\ \\ \\ S =\frac{1}{2-x_1}+\frac{1}{2-x_2}+\frac{1}{2-x_3}+\frac{1}{2-x_4} = \\ \\ = \frac{(2-x_2)(2-x_3)(2-x_4)+(2-x_1)(2-x_3)(2-x_4)+...+(2-x_1)(2-x_2)(2-x_3)}{(2-x_1)(2-x_2)(2-x_3)(2-x_4)} =\\ \\ = \dfrac{f'(2)}{f(2)}\\ \\\\ f(x) = (x-x_1)(x-x_2)(x-x_3)(x-x_4) \\ \\ f'(x) = (x-x_2)(x-x_3)(x-x_4)+(x-x_1)(x-x_3)(x-x_4)+...+\\ +(x-x_1)(x-x_2)(x-x_3)$

Pentru x = 2 avem:

$f(2) = (2-x_1)(2-x_2)(2-x_3)(2-x_4)= 16-12+2+7 = 13 \\\\ f'(2) = 4\cdot 8-12+1 = 21$

$\Rightarrow \boxed{S = \dfrac{21}{13}}$