Bài tập 69 trang 154 SGK Toán 10 NC

a) Điều kiện: \(x \ne  - 1\)

Ta có:

\(\begin{array}{l}
\left| {\frac{{{x^2} - 2}}{{x + 1}}} \right| = 2 \Leftrightarrow \left[ \begin{array}{l}
\frac{{{x^2} - 2}}{{x + 1}} = 2\\
\frac{{{x^2} - 2}}{{x + 1}} =  - 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} - 2 = 2x + 2\\
{x^2} - 2 =  - 2x - 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} - 2x - 4 = 0\\
{x^2} + 2x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1 \pm \sqrt 5 \\
x = 0\\
x =  - 2
\end{array} \right.
\end{array}\)

Vậy \(S = \left\{ {1 \pm \sqrt 5 ;0;2} \right\}\)

b) Điều kiện: \(x \ne 2\)

Ta có:

\(\begin{array}{l}
\left| {\frac{{3x + 4}}{{x - 2}}} \right| \le 3 \Leftrightarrow \left| {3x + 4} \right| \le 3\left| {x - 2} \right|\\
 \Leftrightarrow {\left( {3x + 4} \right)^2} - 9{\left( {x - 2} \right)^2} \le 0\\
 \Leftrightarrow 10\left( {6x - 2} \right) \le 0 \Leftrightarrow x \le \frac{1}{3}
\end{array}\)

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

c) Điều kiện: \(x \ne 3\)

Ta có:

\(\begin{array}{l}
\left| {\frac{{2x - 3}}{{x - 3}}} \right| \ge 1 \Leftrightarrow \left| {2x - 3} \right| \ge \left| {x - 3} \right|\\
 \Leftrightarrow {\left( {2x - 3} \right)^2} - {\left( {x - 3} \right)^2} \ge 0\\
 \Leftrightarrow x\left( {3x - 6} \right) \ge 0 \Leftrightarrow \left[ \begin{array}{l}
x \le 0\\
x \ge 2
\end{array} \right.
\end{array}\)

Vậy \(S = \left( { - \infty ;0} \right] \cup \left[ {2;3} \right) \cup \left[ {3; + \infty } \right)\)

d) Ta có

\(\begin{array}{l}
\left| {2x + 3} \right| = \left| {4 - 3x} \right|\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{2x + 3 = 4 - 3x}\\
{2x + 3 = 3x - 4}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = \frac{1}{5}}\\
{x = 7}
\end{array}} \right.
\end{array}\)

Vậy \(S = \left\{ {\frac{1}{5};7} \right\}\)

-- Mod Toán 10 HỌC247