Bài tập 25 trang 121 SGK Toán 10 NC

Bài tập 25 trang 121 SGK Toán 10 NC

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\frac{{x + 2}}{3} - x + 1 > x + 3\\
 \Leftrightarrow x + 2 - 3x + 3 > 3x + 9
\end{array}\\
{ \Leftrightarrow  - 5x > 4 \Leftrightarrow x <  - \frac{4}{5}}
\end{array}\)

Vậy \(S = \left( { - \infty ; - \frac{4}{5}} \right)\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\frac{{3x + 5}}{2} - 1 \le \frac{{x + 2}}{3} + x\\
 \Leftrightarrow 9x + 15 - 6 \le 2x + 4 + 6x
\end{array}\\
{ \Leftrightarrow x \le  - 5}
\end{array}\)

Vậy \(S = \left( { - \infty ; - 5} \right)\)

c)

\(\begin{array}{l}
\left( {1 - \sqrt 2 } \right)x < 3 - 2\sqrt 2 \\
 \Leftrightarrow \left( {1 - \sqrt 2 } \right)x < {\left( {1 - \sqrt 2 } \right)^2}
\end{array}\)

\( \Leftrightarrow x > \frac{{{{\left( {1 - \sqrt 2 } \right)}^2}}}{{1 - \sqrt 2 }} = 1 - \sqrt 2 \) 

(vì \(1 - \sqrt 2  < 0\))

Vậy \(S = \left( {1 - \sqrt 2 ; + \infty } \right)\)

d)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{\left( {x + \sqrt 3 } \right)^2} \ge {\left( {x - \sqrt 3 } \right)^2} + 2\\
 \Leftrightarrow {\left( {x + \sqrt 3 } \right)^2} - {\left( {x - \sqrt 3 } \right)^2} \ge 2
\end{array}\\
{ \Leftrightarrow 4\sqrt 3 x \ge 2 \Leftrightarrow x \ge \frac{1}{{2\sqrt 3 }}}
\end{array}\)

Vậy \(S = \left[ {\frac{1}{{2\sqrt 3 }}; + \infty } \right)\)

-- Mod Toán 10