Đáp án:
Giải thích các bước giải:
3).
\[\begin{array}{l} 5x - \frac{{x + 1}}{5} - 4 - (2x - 7) < 0 \Leftrightarrow 25x - (x + 1) - 20 - 5(2x - 7) < 0 \Leftrightarrow 14x + 14 < 0 \Leftrightarrow x + 1 < 0 \Leftrightarrow x < - 1\\ S = ( - \infty ; - 1) \end{array}\]
4)Để \[\dfrac{{x - 5}}{{(x + 7)(x - 2)}} > 0\] thì
$\left\{ \begin{array}{l} x - 5 > 0\\ (x + 7)(x - 2) > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x > 5\\ \left[ \begin{array}{l} x > 2\\ x < - 7 \end{array} \right. \end{array} \right. \Leftrightarrow x > 5$ hoặc
$\left\{ \begin{array}{l} x - 5 < 0\\ (x + 7)(x - 2) > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x < 5\\ - 7 < x < 2 \end{array} \right. \Leftrightarrow - 7 < x < 2$
Vậy số nguyên nhỏ nhất để x luôn dương là $x\ge-6\Rightarrow x=-6$
Câu 5:
$\begin{array}{l} \dfrac{{x + 1}}{{x - 1}} - \dfrac{{x + 5}}{{x + 1}} = \dfrac{{{{(x + 1)}^2} - (x + 5)(x - 1)}}{{(x - 1)(x + 1)}} = \dfrac{{ - 2x + 6}}{{(x - 1)(x + 1)}} \ge 0\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} - 2x + 6 \ge 0\\ (x - 1)(x + 1) > 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x < - 1\\ 1 < x \le 3 \end{array} \right.\\ \left\{ \begin{array}{l} - 2x + 6 \le 0\\ (x - 1)(x + 1) < 0 \end{array} \right.(L) \end{array} \right. \end{array}$
Chọn B
