Question

Fie E(x)=($\frac{ x^{2}-4 }{ x^{2} -9}$-1):($\frac{1}{x-3}$+$\frac{1}{x+3}$-$\frac{1}{ x^{2} -9}$),unde x∈R\{-3;$\frac{1}{2}$;3}.
a)Aratati ca E(x) =$\frac{5}{2x-1}$,oricare ar fi x ∈\{-3;$\frac{1}{2}$;3}
b)Determinati numerele reale x pentru care E(x)≤0.

Answer 1

$a) E(x)=( \frac{ x^{2} -4}{ x^{2} -9} - \frac{ x^{2} -9}{ x^{2} -9}):( \frac{x+3}{(x+3)(x-3)}+ \frac{x-3}{(x+3)(x-3)}- \frac{1}{(x+3)(x-3)})= \\ = \frac{ x^{2} -4- x^{2} +9}{ x^{2} -9} : \frac{x+3+x-3-1}{(x+3)(x-3)} = \\ = \frac{5}{(x+3)(x-3)}: \frac{2x-1}{(x+3)(x-3)}= \\ = \frac{5}{(x+3)(x-3)}* \frac{(x+3)(x-3)}{2x-1}= \\ = \frac{5}{2x-1}$

b) Pentru ca E(x) ≤ 0 este suficient ca 2x-1 ≤ 0.

2x-1 ≤ 0 => 2x ≤ 1 => $x \leq \frac{1}{2}$. Deci x∈(-∞;$\frac{1}{2}$).