Đáp án:
\(d)x < 13\)
Giải thích các bước giải:
\(\begin{array}{l}
a)2x\left( {6x - 1} \right) > \left( {3x - 2} \right)\left( {4x - 3} \right)\\
\to 12{x^2} - 2x > 12{x^2} - 17x + 6\\
\to 15x > 6\\
\to x > \frac{6}{{15}}\\
\to x > \frac{2}{5}\\
b){\left( {2x - 1} \right)^2} - 8\left( {x - 1} \right) \le 0\\
\to 4{x^2} - 4x + 1 - 8x + 8 \le 0\\
\to 4{x^2} - 12x + 9 \le 0\\
\to {\left( {2x - 3} \right)^2} \le 0\\
Do:{\left( {2x - 3} \right)^2} \ge 0\forall x \in R\\
\to 2x - 3 = 0\\
\to x = \frac{3}{2}\\
c)(x - 1)(x - 2) > {(x - 1)^2} + 3\\
\to {x^2} - 3x + 2 > {x^2} - 2x + 1 + 3\\
\to x < - 2\\
d)x\left( {2x - 1} \right) - 8 < 5 - 2x\left( {1 - x} \right)\\
\to 2{x^2} - x - 8 < 5 - 2x + 2{x^2}\\
\to x < 13
\end{array}\)
