Question

Fie E(x) = $( \frac{x-1}{x+2} + \frac{x+1}{2-x} + \frac{2+7x}{ x^{2} -4} ) ori ( x+1 + \frac{3-6x}{x+1} )$, x ∈ R/{-2, -1, 2}
a) Aduceţi E(x) la forma cea mai simplă.
b) Determinaţi valorile întregi ale lui x pentru care E(x) ∈ Z.

Answer 1

$E(x)= [\frac{(x-1)(x-2)}{(x+2)(x-2)} - \frac{x+1}{x-2}+ \frac{2+7x}{(x+2)(x-2)} ]*[ \frac{(x+1)^{2}}{x+1} + \frac{3-6x}{x+1}]= \\ = [\frac{(x-1)(x-2)}{(x+2)(x-2)}- \frac{(x+1)(x+2)}{(x+2)(x-2)}+ \frac{2+7x}{(x+2)(x-2)} ] * \frac{(x+1)^2+3-6x}{x+1} = \\ = \frac{(x-1)(x-2)-(x+1)(x+2)+2+7x}{(x+2)(x-2)}* \frac{ x^{2} +2x+1+3-6x}{x+1} = \\ = \frac{ x^{2} -2x-x+2- x^{2} -2x-x-2+2+7x}{(x+2)(x-2)} * \frac{ x^{2} -4x+4}{x+1}= \\ = \frac{x+2}{(x+2)(x-2)}* \frac{(x-2)^2}{x+1} = \\ = \frac{x-2}{x+1} .$

$b)~E(x) \in Z \Leftrightarrow \frac{x-2}{x+1} \in Z \Rightarrow x+1~|~x-2. \\ \\ x+1~|~x-2 \\ x+1~|~x+1 \\ x+1~|~(x+1)-(x-2) \Leftrightarrow x+1~|~3 \Rightarrow (x+1) \in D_{3}=\{-3;-1;1;3\}. \\ \\ x+1=-3 \Rightarrow \boxed{x=-4} .\\ x+1=-1 \Rightarrow x=-2,~nu~convine! \\ x+1=1 \Rightarrow \boxed{x=0}. \\ x+1=3 \Rightarrow x=2,~nu~convine! \\ \\ \underline{Solutie}: \boxed{x \in\{-4;0\}}.$

*Observatie: Am spus ca x=2 si x=-2 nu convin pentru ca ele nu apartin domeniului de definitie. ($x \in R-\{-2;-1;2\}$).

Answer 2

Sper sa te ajute rezolvarea aceasta

a38472dc0d91ed5fc1572e35e543203a.pdf