Решите неравенство

Question 1

Решите неравенство

Question 2

ОДЗ : log₅ (x² -3) ≥ 0

Question 3

$(x^2 - 8x + 7) \cdot \sqrt{ log_{5}(x^2 - 3 )} \le 0$

Область определения $\sqrt{ log_{5}(x^2 - 3 )}$

0\\ log_{5}(x^2 - 3 ) \ge 0\end{cases}" alt="\begin{cases}x^2 - 3>0\\ log_{5}(x^2 - 3 ) \ge 0\end{cases}" align="absmiddle" class="latex-formula">

0\\ log_{5}(x^2 - 3 ) \ge log_51\end{cases}" alt="\begin{cases}(x- \sqrt{3} )(x+ \sqrt{3} )>0\\ log_{5}(x^2 - 3 ) \ge log_51\end{cases}" align="absmiddle" class="latex-formula">

$\begin{cases}x \in (- \infty ;- \sqrt{3} ) \cup ( \sqrt{3};+ \infty )\\ x^2 - 3 \ge1\end{cases}$

$\begin{cases}x \in (- \infty ;- \sqrt{3} ) \cup ( \sqrt{3};+ \infty )\\ x^2 - 4 \ge0\end{cases}$

$\begin{cases}x \in (- \infty ;- \sqrt{3} ) \cup ( \sqrt{3};+ \infty )\\(x-2)(x+2) \ge0\end{cases}$

$\begin{cases}x \in (- \infty ;- \sqrt{3} ) \cup ( \sqrt{3};+ \infty )\\x \in \left(- \infty ;-2 \right] \cup \left[2;+ \infty\right) \end{cases}$

$x \in \left(- \infty ;-2 \right] \cup \left[2;+ \infty\right)$

$x \in \left(- \infty ;-2 \right] \cup \left[2;+ \infty\right) \Rightarrow \sqrt{ log_{5}(x^2 - 3 )} \ge 0$

--------------

$x^2 - 8x + 7 \le 0$

$D=(-8)^2-4 \cdot 1 \cdot 7=64-28=36$

$\sqrt{D}= \sqrt{36} =6$

$x_1= \frac{8-6}{2} = \frac{2}{2}=1$

$x_2= \frac{8+6}{2} = \frac{14}{2}=7$

$x \in \left[ 1;7\right]$

$\begin{cases} x \in \left(- \infty ;-2 \right] \cup \left[2;+ \infty\right)\\ x \in \left[ 1;7\right]\end{cases}$

$x \in \left[ 2;7\right]$

Ответ

$x \in\left\{-2 \right\} \cup \left[ 2;7\right]$

_zn · Answer 1 · 2020-05-23T09:12:37+0000