Đáp án:
$\begin{array}{l}
a)\left( {x - 1} \right)\left( {x + 2} \right) > {\left( {x - 1} \right)^2} + 3\\
\Leftrightarrow {x^2} + 2x - x - 2 > {x^2} - 2x + 1 + 3\\
\Leftrightarrow x + 2x > 4 + 2\\
\Leftrightarrow 3x > 6\\
\Leftrightarrow x > 2\\
Vậy\,x > 2\\
b)x\left( {2x - 1} \right) - 8 < 5 - 2x\left( {1 - x} \right)\\
\Leftrightarrow 2{x^2} - x - 8 < 5 - 2x + 2{x^2}\\
\Leftrightarrow - x + 2x < 5 + 8\\
\Leftrightarrow x < 13\\
Vậy\,x < 13\\
c){\left( {2x + 1} \right)^2} + \left( {1 - x} \right).3x \le {\left( {x + 2} \right)^2}\\
\Leftrightarrow 4{x^2} + 4x + 1 + 3x - 3{x^2} \le {x^2} + 4x + 4\\
\Leftrightarrow 7x - 4x \le 4 - 1\\
\Leftrightarrow 3x \le 3\\
\Leftrightarrow x \le 1\\
Vậy\,x \le 1\\
d)\left( {x - 4} \right)\left( {x + 4} \right) \ge {\left( {x + 3} \right)^2} + 5\\
\Leftrightarrow {x^2} - 16 \ge {x^2} + 6x + 9 + 5\\
\Leftrightarrow 6x \le - 16 - 14\\
\Leftrightarrow 6x \le - 30\\
\Leftrightarrow x \le - 5\\
Vậy\,x \le - 5\\
B2)a){\left( {x - 3} \right)^2} < {x^2} - 5x + 4\\
\Leftrightarrow {x^2} - 6x + 9 < {x^2} - 5x + 4\\
\Leftrightarrow - 5x + 6x > 9 - 4\\
\Leftrightarrow x > 5\\
Vậy\,x > 5\\
b)\left( {x - 3} \right)\left( {x + 3} \right) \le {\left( {x + 2} \right)^2} + 3\\
\Leftrightarrow {x^2} - 9 \le {x^2} + 4x + 4 + 3\\
\Leftrightarrow 4x \ge - 9 - 7\\
\Leftrightarrow x \ge \dfrac{{ - 16}}{4}\\
\Leftrightarrow x \ge - 4\\
Vậy\,x \ge - 4
\end{array}$
