Đáp án:
b) \(x \le - 6\)
Giải thích các bước giải:
\(\begin{array}{l}
a)4x - 8 \ge 3\left( {3x - 1} \right) - 2x + 1\\
\to 4x - 8 \ge 9x - 3 - 2x + 1\\
\to 3x \le - 6\\
\to x \le - 2\\
b)\left( {x + 3} \right)\left( {x + 2} \right) + {\left( {x + 4} \right)^2} \le 2x\left( {x + 5} \right) + 4\\
\to {x^2} + 5x + 6 + {x^2} + 8x + 16 \le 2{x^2} + 10x + 4\\
\to 3x \le - 18\\
\to x \le - 6\\
c)\left( {x - 2} \right)\left( {2x - 3} \right) + 3\left( {x + 1} \right) < 2{\left( {x - 1} \right)^2} - 4x\\
\to 2{x^2} - 7x + 6 + 3x + 3 < 2\left( {{x^2} - 2x + 1} \right) - 4x\\
\to 2{x^2} - 4x + 9 < 2{x^2} - 8x + 2\\
\to 4x < - 7\\
\to x < - \dfrac{7}{4}
\end{array}\)
