Question

va rog daca ma puteți ajută la ex 4 imi trebuie urgent!​

Answer 1

$E = \dfrac{x_1^2+3x_1+2}{x_1^2+5x_1+2}+\dfrac{x_2^2+3x_2+2}{x_2^2+5x_2+2} \\ \\ \\2x^2-7x+2 = 0\\ \\x^2+3x+2 = -x^2+10x\\ x^2+5x+2 = -x^2+12x \\ \\ \Rightarrow E = \dfrac{-x_1^2+10x_1}{-x_1^2+12x_1}+\dfrac{-x_2^2+10x_2}{-x_2^2+12x_2} =\\ \\ = \dfrac{-x_1(x_1-10)}{-x_1(x_1-12)}+\dfrac{-x_2(x_2-10)}{-x_2(x_2-12)}= \\ \\ = \dfrac{x_1-10}{x_1-12}+\dfrac{x_2-10}{x_2-12} = \dfrac{x_1-12+2}{x_1-12}+\dfrac{x_2-12+2}{x_2-12} =$

$\\ =1+1+\dfrac{2}{x_1-12}+\dfrac{2}{x_2-12} = 2+\dfrac{2(x_2-12)+2(x_1-12)}{(x_1-12)(x_2-12)} = \\ \\ = 2+\dfrac{2(x_1+x_2)-48}{\dfrac{f(12)}{2}} =2+\dfrac{7-48}{\dfrac{2\cdot 12^2-7\cdot 12+2}{2}} = \\ \\ \\= 2-\dfrac{41}{103} = \boxed{\dfrac{165}{103}}\\ \\ \\f(x) = 2(x-x_1)(x-x_2) = 2(x_1-x)(x_2-x) \\ f(12) = 2(x_1-12)(x_2-12) \Rightarrow (x_1-12)(x_2-12) = \dfrac{f(12)}{2}$