Bài tập 62 trang 146 SGK Toán 10 NC

a) Ta có:

\(\begin{array}{l}
\left\{ {\begin{array}{*{20}{l}}
{4x - 3 < 3x + 4}\\
{{x^2} - 7x + 10 \le 0}
\end{array}} \right.\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x < 7}\\
{2 \le x \le 5}
\end{array}} \right. \Leftrightarrow 2 \le x \le 5
\end{array}\)

Vậy S = [2;5]

b) Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
2{x^2} + 9x - 7 > 0\\
{x^2} + x - 6 \le 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x < \frac{{ - 9 - \sqrt {137} }}{4}\\
x > \frac{{ - 9 + \sqrt {137} }}{4}
\end{array} \right.\\
 - 3 \le x \le 2
\end{array} \right.\\
 \Leftrightarrow \frac{{ - 9 + \sqrt {137} }}{4} < x \le 2
\end{array}\)

Vậy tập nghiệm là \(S = \left( {\frac{{ - 9 + \sqrt {137} }}{4};2} \right]\)

c) Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 9 < 0\\
\left( {x - 1} \right)\left( {3{x^2} + 7x + 4} \right) \ge 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
 - 3 < x < 3\\
\left[ \begin{array}{l}
 - \frac{4}{3} \le x \le  - 1\\
x \ge 1
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
 - \frac{4}{3} \le x \le  - 1\\
1 \le x < 3
\end{array} \right.
\end{array}\)

Vậy \(S = \left[ { - \frac{4}{3}; - 1} \right] \cup \left[ {1;3} \right)\)

-- Mod Toán 10 HỌC247