Giải bất phương trình sau: a, (x-3)(√2 -x)>0 b, (x+2)(x^2-4)<=0 c,. x2+4x+3<=0 d,. (1-3x)(-6x2 + 5x + 1)>= 0 e,. 9×2 – 4x <= 0 f,. -6x2 + x + 1 >=

Đáp án:

$\begin{array}{l}
a)\left( {x – 3} \right)\left( {\sqrt 2  – x} \right) > 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 3 > 0\\
\sqrt 2  – x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 3 < 0\\
\sqrt 2  – x < 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 3\\
x < \sqrt 2 
\end{array} \right.\left( {ktm} \right)\\
\left\{ \begin{array}{l}
x < 3\\
x > \sqrt 2 
\end{array} \right.
\end{array} \right.\\
Vậy\,\sqrt 2  < x < 3\\
b)\left( {x + 2} \right)\left( {{x^2} – 4} \right) \le 0\\
 \Leftrightarrow \left( {x + 2} \right)\left( {x + 2} \right)\left( {x – 2} \right) \le 0\\
 \Leftrightarrow {\left( {x + 2} \right)^2}\left( {x – 2} \right) \le 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x – 2 \le 0\\
x + 2 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x \le 2\\
x =  – 2
\end{array} \right.\\
 \Leftrightarrow x \le 2\\
Vậy\,x \le 2\\
c){x^2} + 4x + 3 \le 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x + 3} \right) \le 0\\
 \Leftrightarrow  – 3 \le x \le  – 1\\
Vậy\, – 3 \le x \le  – 1\\
d)\left( {1 – 3x} \right)\left( { – 6{x^2} + 5x + 1} \right) \ge 0\\
 \Leftrightarrow \left( {3x – 1} \right)\left( {6{x^2} – 5x – 1} \right) \ge 0\\
 \Leftrightarrow \left( {3x – 1} \right)\left( {6x + 1} \right)\left( {x – 1} \right) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
 – \dfrac{1}{6} \le x \le \dfrac{1}{3}
\end{array} \right.\\
Vậy\, – \dfrac{1}{6} \le x \le \dfrac{1}{3}\,hoặc\,x \ge 1\\
e)9{x^2} – 4x \le 0\\
 \Leftrightarrow x\left( {9x – 4} \right) \le 0\\
 \Leftrightarrow 0 \le x \le \dfrac{4}{9}\\
Vậy\,0 \le x \le \dfrac{4}{9}\\
f) – 6{x^2} + x + 1 \ge 0\\
 \Leftrightarrow 6{x^2} – x – 1 \ge 0\\
 \Leftrightarrow \left( {2x – 1} \right)\left( {3x + 1} \right) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x \le  – \dfrac{1}{3}
\end{array} \right.\\
Vậy\,x \le  – \dfrac{1}{3}\,hoặc\,x \ge \dfrac{1}{2}
\end{array}$