Đáp án:
B2:
\(x \in \left( { - 1;0} \right] \cup \left( {1; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
1)DK:x \ne \pm 1\\
\dfrac{{3x{{\left( {x + 1} \right)}^2} + 2x{{\left( {x - 1} \right)}^2} - 4x\left( {{x^2} + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} \ge 0\\
\to \dfrac{{3x\left( {{x^2} + 2x + 1} \right) + 2x\left( {{x^2} - 2x + 1} \right) - 4{x^3} - 4x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} \ge 0\\
\to \dfrac{{3{x^3} + 6{x^2} + 3x + 2{x^3} - 4{x^2} + 2x - 4{x^3} - 4x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} \ge 0\\
\to \dfrac{{{x^3} + 2{x^2} + x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} \ge 0\\
\to \dfrac{{x\left( {{x^2} + 2x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} \ge 0\\
\to \dfrac{{x{{\left( {x + 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} \ge 0\\
\to \dfrac{{x\left( {x + 1} \right)}}{{x - 1}} \ge 0
\end{array}\)
BXD:
x -∞ -1 0 1 +∞
f(x) - // + 0 - // +
\(KL:x \in \left( { - 1;0} \right] \cup \left( {1; + \infty } \right)\)
\(\begin{array}{l}
B3:\\
2)\left( {{x^2} + x} \right)\left( {{x^2} + 5x + 6} \right) - 24 > 0\\
\to {x^4} + 5{x^3} + 6{x^2} + {x^3} + 5{x^2} + 6x - 24 > 0\\
\to {x^4} + 6{x^3} + 11{x^2} + 6x - 24 > 0\\
\to {x^4} - {x^3} + 7{x^3} - 7{x^2} + 18x - 18x + 24x - 24 > 0\\
\to {x^3}\left( {x - 1} \right) + 7{x^2}\left( {x - 1} \right) + 18x\left( {x - 1} \right) + 24\left( {x - 1} \right) > 0\\
\to \left( {x - 1} \right)\left( {{x^3} + 7{x^2} + 18x + 24} \right) > 0\\
\to \left( {x - 1} \right)\left( {x + 4} \right)\left( {{x^2} + 3x + 6} \right) > 0\\
\to \left( {x - 1} \right)\left( {x + 4} \right) > 0\left( {do:{x^2} + 3x + 6 > 0\forall x} \right)
\end{array}\)
BXD:
x -∞ -4 1 +∞
f(x) + 0 - 0 +
\(KL:x \in \left( { - \infty ; - 4} \right) \cup \left( {1; + \infty } \right)\)
