Bài tập 70 trang 154 SGK Toán 10 NC

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 HỌC247