Đáp án:
$\begin{array}{l}
1)a)\dfrac{{3x - 2}}{5} \ge \dfrac{x}{2} + 0,8\\
\Rightarrow \dfrac{3}{5}.x - \dfrac{2}{5} \ge \dfrac{x}{2} + \dfrac{4}{5}\\
\Rightarrow \dfrac{3}{5}.x - \dfrac{x}{2} \ge \dfrac{4}{5} + \dfrac{2}{5}\\
\Rightarrow \dfrac{1}{{10}}.x \ge \dfrac{6}{5}\\
\Rightarrow x \ge 12\\
Khi:1 - \dfrac{{2x - 5}}{6} > \dfrac{{3 - x}}{4}\\
\Rightarrow 1 - \dfrac{x}{3} + \dfrac{5}{6} > \dfrac{3}{4} - \dfrac{x}{4}\\
\Rightarrow \dfrac{x}{3} - \dfrac{x}{4} < 1 + \dfrac{5}{6} - \dfrac{3}{4}\\
\Rightarrow \dfrac{x}{{12}} < \dfrac{{13}}{{12}}\\
\Rightarrow x < 13\\
\Rightarrow 12 \le x < 13\\
\Leftrightarrow x = 12\\
Vậy\,x = 12\\
b)2\left( {3x - 4} \right) < 3\left( {4x - 3} \right) + 16\\
\Leftrightarrow 6x - 8 < 12x - 9 + 16\\
\Leftrightarrow 12x - 6x > - 15\\
\Rightarrow x > \dfrac{{ - 5}}{2}\\
a.\left( {1 + x} \right) < 3x + 5\\
\Rightarrow a + a.x < 3x + 5\\
\Rightarrow \left( {a - 3} \right).x < 5 - a\\
B2)\\
a)Dkxd:x \ne 1;x \ne - 1\\
A = \left( {\dfrac{1}{{1 - x}} + \dfrac{2}{{x + 1}} - \dfrac{{5 - x}}{{1 - {x^2}}}} \right):\dfrac{{1 - 2x}}{{{x^2} - 1}}\\
= \dfrac{{ - x - 1 + 2\left( {x - 1} \right) + 5 - x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{1 - 2x}}\\
= \dfrac{{ - x - 1 + 2x - 2 + 5 - x}}{{1 - 2x}}\\
= \dfrac{2}{{1 - 2x}}\\
b)x \ne 1;x \ne - 1\\
A > 0\\
\Leftrightarrow \dfrac{2}{{1 - 2x}} > 0\\
\Leftrightarrow 1 - 2x > 0\\
\Leftrightarrow x < \dfrac{1}{2}\\
Vậy\,x < \dfrac{1}{2};x \ne - 1
\end{array}$
