a, 

$x^2+x+1=(x+\dfrac{1}{2})^2+\dfrac{3}{4}>0\forall x$

Đặt $t=x^2+x+1$

$\Rightarrow t>0$

$t-1<\dfrac{42}{t}$

$\Leftrightarrow t^2-t-42<0$

$\Leftrightarrow -6<t<7$

$x^2+x+1>-6\Leftrightarrow x^2+x+7>0$ (luôn đúng)

$x^2+x+1<7\Leftrightarrow x^2+x-6<0\Leftrightarrow -3<x<2$ 

b,

$\sqrt{2x^2+3x+1}\le -2x-1$ (*)

ĐK: $2x^2+3x+1\ge 0, -2x-1\ge 0$

$\Leftrightarrow x\le -1$ hoặc $x\ge -0,5$; $x\le 0,5$

$\Rightarrow x\le -1$ hoặc $-0,5\le x\le 0,5$

(*)$\Leftrightarrow 2x^2+3x+1<4x^2+4x+1$

$\Leftrightarrow 2x^2+x>0$

$\Leftrightarrow x<\dfrac{-1}{2}; x>0$

$\to x\le -1; 0<x\le 0,5$

Đáp án:

b. \(x \in \left( { - \infty ; - 1} \right] \cup \left\{ { - \frac{1}{2}} \right\}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.{x^2} + x < \dfrac{{42}}{{{x^2} + x + 1}}\left( 1 \right)\\
Đặt:{x^2} + x + 1 = t\left( {t > 0} \right)\\
\left( 1 \right) \to t - 1 < \dfrac{{42}}{t}\\
 \to \dfrac{{{t^2} - t - 42}}{t} < 0\\
 \to \dfrac{{\left( {t - 7} \right)\left( {t + 6} \right)}}{t} < 0\\
 \to \dfrac{{t - 7}}{t} < 0\left( {do:t + 6 > 0\forall t > 0} \right)\\
 \to t - 7 < 0\\
 \to 0 < t < 7\\
 \to \left\{ \begin{array}{l}
{x^2} + x + 1 > 0\left( {ld} \right)\forall x \in R\\
{x^2} + x + 1 < 7
\end{array} \right.\\
 \to \left( {x - 2} \right)\left( {x + 3} \right) < 0\\
 \to x \in \left( { - 3;2} \right)\\
b.\sqrt {2{x^2} + 3x + 1}  \le  - 1 - 2x\\
 \to \left\{ \begin{array}{l}
2{x^2} + 3x + 1 \ge 0\\
 - 1 - 2x \ge 0\\
2{x^2} + 3x + 1 \le 1 + 4x + 4{x^2}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left( {2x + 1} \right)\left( {x + 1} \right) \ge 0\\
x \le  - \dfrac{1}{2}\\
2{x^2} + x \ge 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x \in \left( { - \infty ; - 1} \right] \cup \left[ { - \dfrac{1}{2}; + \infty } \right)\\
x \le  - \dfrac{1}{2}\\
x\left( {2x + 1} \right) \ge 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x \in \left( { - \infty ; - 1} \right] \cup \left[ { - \dfrac{1}{2}; + \infty } \right)\\
x \le  - \dfrac{1}{2}\\
x \in \left( { - \infty ; - \dfrac{1}{2}} \right] \cup \left[ {0; + \infty } \right)
\end{array} \right.\\
 \to x \in \left( { - \infty ; - 1} \right] \cup \left\{ { - \dfrac{1}{2}} \right\}
\end{array}\)