Question

Rezolvați ecuațiile:
a )|x+6|>2
b)log x+2(x^2-6x+8)=2.

Answer 1

$\displaystyle a) |x+6|\ \textgreater \ 2 \\ \\ x+6\ \textless \ -2 \Rightarrow x\ \textless \ -2-6 \Rightarrow x\ \textless \ -8\Rightarrow x \in (-\infty,-8) \\ \\ x+6\ \textgreater \ 2 \Rightarrow x\ \textgreater \ 2-6 \Rightarrow x\ \textgreater \ -4 \Rightarrow x \in (-4, \infty) \\ \\ x \in (-\infty,-8) \cup (-4, \infty)$

$\displaystyle b)log_{x+2}\left(x^2-6x+8\right)=2~~~~~~~~~~~~~~~~~~~~~~~~~~\left\{\begin{array}{ccc}x+2\ \textgreater \ 0\\x+2 \not =1\\x^2-6x+8\ \textgreater \ 0\end{array}\right\\ \\x^2-6x+8=(x+2)^2\\ \\ x^2-6x+8=x^2+4x+4\\ \\x^2-6x+8-x^2-4x-4=0\\ \\-10x+4=0\\ \\-10x=-4\\ \\ x=\frac{4}{10}\\ \\x=\frac{2}{5}\\ \\x=\frac{2}{5} \Rightarrow \left\{\begin{array}{ccc} \frac{2}{5}+2\ \textgreater \ 0~A\\ \frac{2}{5}+2\not=1~A\\\left( \frac{2}{5}\right)^2-6 \cdot \frac{2}{5}+8\ \textgreater \ 0~A\end{array}\right\Rightarrow x= \frac{2}{5}~este~solutie~a~ecuatiei$
$\displaystyle S=\left\{ \frac{2}{5} \right\}$