Bài tập 29 trang 121 SGK Toán 10 NC

Bài tập 29 trang 121 SGK Toán 10 NC

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\left\{ {\begin{array}{*{20}{l}}
{\frac{{5x + 2}}{3} \ge 4 - x}\\
{\frac{{6 - 5x}}{{13}} < 3x + 1}
\end{array}} \right.\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{5x + 2 \ge 12 - 3x}\\
{6 - 5x < 39x + 13}
\end{array}} \right.
\end{array}\\
\begin{array}{l}
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{8x \ge 10}\\
{44x >  - 7}
\end{array}} \right.\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x \ge \frac{5}{4}}\\
{x >  - \frac{7}{{44}}}
\end{array}} \right. \Leftrightarrow x \ge \frac{5}{4}
\end{array}
\end{array}\)

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

b)

\(\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{{{\left( {1 - x} \right)}^2} > 5 + 3x + {x^2}}\\
{{{\left( {x + 2} \right)}^2} < {x^3} + 6{x^2} - 7x - 5}
\end{array}} \right.}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{1 - 2x + {x^2} > 5 + 3x + {x^2}}\\
\begin{array}{l}
{x^3} + 6{x^2} + 12x + 8 < {x^3}\\
\,\,\, + 6{x^2} - 7x - 5
\end{array}
\end{array}} \right.}\\
\begin{array}{l}
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{5x <  - 4}\\
{19x <  - 13}
\end{array}} \right.\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x <  - \frac{4}{5}}\\
{x <  - \frac{{13}}{{19}}}
\end{array}} \right. \Leftrightarrow x <  - \frac{4}{5}
\end{array}
\end{array}\)

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

c)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\left\{ {\begin{array}{*{20}{l}}
{\frac{{4x - 5}}{7} < x + 3}\\
{\frac{{3x + 8}}{4} > 2x - 5}
\end{array}} \right.\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{4x - 5 < 7x + 21}\\
{3x + 8 > 8x - 20}
\end{array}} \right.
\end{array}\\
\begin{array}{l}
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{3x >  - 26}\\
{5x < 28}
\end{array}} \right.\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x >  - \frac{{26}}{3}}\\
{x < \frac{{28}}{5}}
\end{array}} \right.\\
 \Leftrightarrow  - \frac{{26}}{3} < x < \frac{{28}}{5}
\end{array}
\end{array}\)

Vậy \(S = \left( { - \frac{{26}}{3};\frac{{28}}{5}} \right)\)

d)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\left\{ {\begin{array}{*{20}{l}}
{x - 1 \le 2x - 3}\\
{3x < x + 5}\\
{\frac{{5 - 3x}}{2} \le x - 3}
\end{array}} \right.\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x \ge 2}\\
{2x < 5}\\
{5 - 3x \le 2x - 6}
\end{array}} \right.
\end{array}\\
\begin{array}{l}
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x \ge 2}\\
{x < \frac{5}{2}}\\
{x \ge \frac{{11}}{5}}
\end{array}} \right.\\
 \Leftrightarrow \frac{{11}}{5} \le x < \frac{5}{2}
\end{array}
\end{array}\)

Vậy \(S = \left[ {\frac{{11}}{5};\frac{5}{2}} \right)\)

-- Mod Toán 10