Bài tập 67 trang 151 SGK Toán 10 NC

a) Ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sqrt {{x^2} + x - 6}  < x - 1\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{{x^2} + x - 6 \ge 0}\\
{x - 1 > 0}\\
{{x^2} + x - 6 < {{\left( {x - 1} \right)}^2}}
\end{array}} \right.
\end{array}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{\left[ {\begin{array}{*{20}{l}}
{x \le 3}\\
{x \ge 2}
\end{array}} \right.}\\
{x > 1}\\
{3x < 7}
\end{array}} \right. \Leftrightarrow 2 \le x < \frac{7}{3}}
\end{array}\)

Vậy \(S = \left[ {2;\frac{7}{3}} \right)\)

b) Ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sqrt {2x - 1}  \le 2x - 3\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{2x - 1 \ge 0}\\
{2x - 3 \ge 0}\\
{2x - 1 \le {{\left( {2x - 3} \right)}^2}}
\end{array}} \right.
\end{array}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x \ge \frac{1}{2}}\\
{x \ge \frac{3}{2}}\\
{4{x^2} - 14x + 10 \ge 0}
\end{array}} \right.}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x \ge \frac{3}{2}}\\
{\left[ {\begin{array}{*{20}{l}}
{x \le 1}\\
{x \ge \frac{5}{2}}
\end{array}} \right.}
\end{array}} \right. \Leftrightarrow x \ge \frac{5}{2}}
\end{array}\)

Vậy \(S = \left[ {\frac{5}{2}; + \infty } \right)\)

c) Ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sqrt {2{x^2} - 1}  > 1 - x\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{1 - x < 0}\\
{2{x^2} - 1 \ge 0}
\end{array}} \right.}\\
{\left\{ {\begin{array}{*{20}{l}}
{1 - x \ge 0}\\
{2{x^2} - 1 > {{\left( {1 - x} \right)}^2}}
\end{array}} \right.}
\end{array}} \right.
\end{array}\\
\begin{array}{l}
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x > 1}\\
{\left\{ {\begin{array}{*{20}{l}}
{x \le 1}\\
{{x^2} + 2x - 2 > 0}
\end{array}} \right.}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x > 1}\\
{\left\{ {\begin{array}{*{20}{l}}
{x \le 1}\\
{\left[ {\begin{array}{*{20}{l}}
{x <  - 1 - \sqrt 3 }\\
{x >  - 1 + \sqrt 3 }
\end{array}} \right.}
\end{array}} \right.}
\end{array}} \right.
\end{array}\\
{ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x <  - 1 - \sqrt 3 }\\
{x >  - 1 + \sqrt 3 }
\end{array}} \right.}
\end{array}\)

Vậy \(S = \left( { - \infty ; - 1 - \sqrt 3 } \right) \cup \left( { - 1 + \sqrt 3 ; + \infty } \right)\)

d) Ta có:

\(\begin{array}{l}
\sqrt {{x^2} - 5x - 14}  \ge 2x - 1\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x - 1 < 0\\
{x^2} - 5x - 14 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - 1 \ge 0\\
{x^2} - 5x - 14 \ge {\left( {2x - 1} \right)^2}
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x < \frac{1}{2}\\
\left[ \begin{array}{l}
x \le  - 2\\
x \ge 7
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge \frac{1}{2}\\
3{x^2} + x + 15 \le 0
\end{array} \right.
\end{array} \right. \Leftrightarrow x \le  - 2
\end{array}\)

Vậy \(S = \left( { - \infty ; - 2} \right]\)

-- Mod Toán 10 HỌC247