Giải các bất phương trình sau:
a) |x2 – 5x + 4| ≤ x2 + 6x + 5
b) 4x2 + 4x - |2x + 1| ≥ 5
a) |x2 – 5x + 4| ≤ x2 + 6x + 5
⇔ - x2 – 6x – 5 ≤ x2 – 5x + 4 ≤ x2 + 6x + 5
\( \Leftrightarrow \left\{ \begin{array}{l}
2{x^2} + x + 9 \ge 0\\
11x \ge - 1
\end{array} \right. \)
\(\Leftrightarrow x \ge - \frac{1}{{11}}\)
Vậy \(S = \left[ { - \frac{1}{{11}}; + \infty } \right)\)
b) Ta có: 4x2 + 4x - |2x + 1| ≥ 5
⇔ |2x + 1| ≤ 4x2 + 4x – 5
⇔ - 4x2 – 4x + 5 ≤ 2x + 1 ≤ 4x2 + 4x – 5
\(\begin{array}{l}
\Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{4{x^2} + 6x - 4 \ge 0}\\
{4{x^2} + 2x - 6 \ge 0}
\end{array}} \right.\\
\Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{\left[ {\begin{array}{*{20}{l}}
{x \le - 2}\\
{x \ge \frac{1}{2}}
\end{array}} \right.}\\
{\left[ {\begin{array}{*{20}{l}}
{x \le - \frac{3}{2}}\\
{x \ge 1}
\end{array}} \right.}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x \le - 2}\\
{x \ge 1}
\end{array}} \right.
\end{array}\)
Vậy \(S = \left( { - \infty ; - 2} \right] \cup \left[ {1; + \infty } \right)\)
-- Mod Toán 10
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