giải các hệ bất phương trình sau
1.5x-10>0
x^2-x-12<0
2.4x-7-x^2<0
x^2-2x-3≥0
3.3x^2-20x-7<0
2x^2-13x+18>0
4.x^2+x+5<0
x^2-6x-27>0
giải các hệ bất phương trình sau
1.5x-10>0
x^2-x-12<0
2.4x-7-x^2<0
x^2-2x-3≥0
3.3x^2-20x-7<0
2x^2-13x+18>0
4.x^2+x+5<0
x^2-6x-27>0
Đáp án:
$\begin{array}{l}
1)\left\{ \begin{array}{l}
5x – 10 > 0\\
{x^2} – x – 12 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x > 2\\
\left( {x – 4} \right)\left( {x + 3} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x > 2\\
– 3 < x < 4
\end{array} \right.\\
\Leftrightarrow 2 < x < 4\\
Vậy\,2 < x < 4\\
2)\left\{ \begin{array}{l}
4x – 7 – {x^2} < 0\\
{x^2} – 2x – 3 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} – 4x + 7 > 0\\
\left( {x – 3} \right)\left( {x + 1} \right) \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x – 2} \right)^2} + 3 > 0\left( {tm} \right)\\
\left[ \begin{array}{l}
x \ge 3\\
x \le – 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le – 1
\end{array} \right.\\
Vậy\,x \ge 3\,hoặc\,x \le – 1\\
3)\left\{ \begin{array}{l}
3{x^2} – 20x – 7 < 0\\
2{x^2} – 13x + 18 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {3x + 1} \right)\left( {x – 7} \right) < 0\\
\left( {2x – 9} \right)\left( {x – 2} \right) > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\frac{{ – 1}}{3} < x < 7\\
\left[ \begin{array}{l}
x > \frac{9}{2}\\
x < 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
– \frac{1}{3} < x < 2\\
\frac{9}{2} < x < 7
\end{array} \right.\\
4)\left\{ \begin{array}{l}
{x^2} + x + 5 < 0\\
{x^2} – 6x – 27 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + \frac{1}{2}} \right)^2} + \frac{{19}}{4} < 0\left( {ktm} \right)\\
{x^2} – 6x – 27 > 0
\end{array} \right.
\end{array}$
Vậy hệ pt vô nghiệm