Đáp án:
$\begin{array}{l}
1)m\left( {x - m} \right) \le 4x + 5\\
\Rightarrow mx - {m^2} \le 4x + 5\\
\Rightarrow \left( {m - 4} \right).x \le {m^2} + 5\\
+ Khi:m \ge 4 \Rightarrow x \le \dfrac{{{m^2} + 5}}{{m - 4}}\\
+ Khi:m < 4 \Rightarrow x \ge \dfrac{{{m^2} + 5}}{{m - 4}}\\
2)\left( {x + 1} \right).k + x < 3x + 4\\
\Rightarrow \left( {k + 1 - 3} \right).x < 4 - k\\
\Rightarrow \left( {k - 2} \right).x < 4 - k\\
+ Khi:k \ge 2 \Rightarrow x < \dfrac{{4 - k}}{{k - 2}}\\
+ Khi:k < 2 \Rightarrow x > \dfrac{{4 - k}}{{k - 2}}\\
3)\left( {a + 1} \right).x + a + 3 \ge 4x + 1\\
\Rightarrow \left( {a + 1 - 4} \right).x \ge - a - 3 + 1\\
\Rightarrow \left( {a - 3} \right).x \ge - a - 2\\
+ Khi:a \ge 3 \Rightarrow x \ge \dfrac{{ - a - 2}}{{a - 3}}\\
+ Khi:a < 3 \Rightarrow x \le \dfrac{{ - a - 2}}{{a - 3}}\\
4)m\left( {x - m} \right) > 2\left( {4 - x} \right)\\
\Rightarrow \left( {m + 2} \right).x > {m^2} + 8\\
+ Khi:m \ge - 2 \Rightarrow x > \dfrac{{{m^2} + 8}}{{m + 2}}\\
+ Khi:m < - 2 \Rightarrow x < \dfrac{{{m^2} + 8}}{{m + 2}}\\
5)3x + {m^2} > m\left( {x + 3} \right)\\
\Rightarrow \left( {m - 3} \right).x < {m^2} - 3m\\
+ Khi:m = 3\left( {ktm} \right)\\
+ Khi:m > 3 \Rightarrow x < m\\
+ Khi:m < 3 \Rightarrow x > m
\end{array}$
