Bài tập 73 trang 154 SGK Toán 10 NC

a) Ta có:

\(\begin{array}{l}
\sqrt {{x^2} - x - 12}  \ge x - 1\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 1 < 0\\
{x^2} - x - 12 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 \ge 0\\
{x^2} - x - 12 \ge {\left( {x - 1} \right)^2}
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x < 1\\
\left[ \begin{array}{l}
x \le  - 3\\
x \ge 4
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge 1\\
x \ge 13
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le  - 3\\
x \ge 13
\end{array} \right.
\end{array}\)

b) Ta có:

\(\begin{array}{l}
\sqrt {{x^2} - 4x - 12}  > 2x + 3\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x + 3 < 0\\
{x^2} - 4x - 12 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - 3 \ge 0\\
{x^2} - 4x - 12 > {\left( {2x + 3} \right)^2}
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x <  - \frac{3}{2}\\
\left[ \begin{array}{l}
x \le  - 2\\
x \ge 6
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge \frac{3}{2}\\
3{x^2} + 16x + 21 < 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x \le  - 2\\
\left\{ \begin{array}{l}
x \ge \frac{3}{2}\\
 - 3 < x <  - \frac{7}{3}
\end{array} \right.
\end{array} \right. \Leftrightarrow x \le  - 2
\end{array}\)

c) Ta có \(\frac{{\sqrt {x + 5} }}{{1 - x}} < 1 \)

\(\begin{array}{l}
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{1 - x > 0}\\
{\sqrt {x + 5}  < 1 - x}
\end{array}} \right.{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( I \right)}\\
{\left\{ {\begin{array}{*{20}{l}}
{1 - x < 0}\\
{\sqrt {x + 5}  > 1 - x}
\end{array}} \right.{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {II} \right)}
\end{array}} \right.\\
\begin{array}{*{20}{l}}
{\left( I \right) \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x < 1}\\
{x + 5 \ge 0}\\
{x + 5 < {{\left( {1 - x} \right)}^2}}\\
{ - 5 \le x < 1}
\end{array}} \right.}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x < 1}\\
{x \ge  - 5}\\
{{x^2} - 3x - 4 > 0}
\end{array}} \right.}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x < 1}\\
{x \ge  - 5}\\
{{x^2} - 3x - 4 > 0}
\end{array}} \right.}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{ - 5 \le x < 1}\\
{\left[ {\begin{array}{*{20}{l}}
{x <  - 1}\\
{x > 4}
\end{array}} \right.}
\end{array}} \right. \Leftrightarrow  - 5 \le x < 1}
\end{array}\\
\left( {II} \right) \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x > 1}\\
{\left[ {\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{1 - x < 0}\\
{x + 5 \ge 0}
\end{array}} \right.}\\
{\left\{ {\begin{array}{*{20}{l}}
{1 - x \ge 0}\\
{x + 5 > {{\left( {1 - x} \right)}^2}}
\end{array}} \right.}
\end{array}} \right.}
\end{array}} \right.\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x > 1}\\
{1 - x < 0}\\
{x + 5 \ge 0}
\end{array}} \right. \Leftrightarrow x > 1
\end{array}\)

Vậy \(S = \left[ { - 5; - 1} \right) \cup \left( {1; + \infty } \right)\)

-- Mod Toán 10 HỌC247