Giải các bất phương trình sau:
a) \(\left( {x + 1} \right)\left( {2x - 1} \right) + x \le 3 + 2{x^2}\)
b) \(\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) - x > {x^3} + 6{x^2} - 5\)
c) \(x + \sqrt x > \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)\)
a) \(\left( {x + 1} \right)\left( {2x - 1} \right) + x \le 3 + 2{x^2}\)
\(\begin{array}{l}
\Leftrightarrow 2{x^2} + 2x - 1 \le 3 + 2{x^2}\\
\Leftrightarrow 2x \le 4 \Leftrightarrow x \le 2
\end{array}\)
b) \(\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) - x > {x^3} + 6{x^2} - 5\)
\(\begin{array}{l}
\Leftrightarrow {x^3} + 6{x^2} + 10x + 6 > {x^3} + 6{x^2} - 5\\
\Leftrightarrow 10x > - 11 \Leftrightarrow x > - \frac{{11}}{{10}}
\end{array}\)
c) \(x + \sqrt x > \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)\)
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x + \sqrt x > 2x + \sqrt x - 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
3 > x
\end{array} \right. \Leftrightarrow 0 \le x < 3
\end{array}\)
-- Mod Toán 10