Đáp án:
c) \(x \ge \dfrac{2}{{11}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a){x^2} - 3x + 1 > 2\left( {x - 1} \right) - x\left( {3 - x} \right)\\
\to {x^2} - 3x + 1 > 2x - 2 + {x^2} - 3x\\
\to 2x < 3\\
\to x < \dfrac{3}{2}\\
b){\left( {x - 1} \right)^2} + {x^2} \le {\left( {x + 1} \right)^2} + {\left( {x + 2} \right)^2}\\
\to {x^2} - 2x + 1 + {x^2} \le {x^2} + 2x + 1 + {x^2} + 4x + 4\\
\to 8x \ge - 4\\
\to x \ge - \dfrac{1}{2}\\
c)\left( {{x^2} + 1} \right)\left( {x - 6} \right) \le {\left( {x - 2} \right)^3}\\
\to {x^3} - 6{x^2} + x - 6 \le {x^3} - 6{x^2} + 12x - 8\\
\to 11x \ge 2\\
\to x \ge \dfrac{2}{{11}}
\end{array}\)
