Đáp án:
Câu 15: C
Câu 16: A
Giải thích các bước giải:
\(\begin{array}{l}
C15:DK:x \ne \pm 1\\
\dfrac{1}{{x + 1}} < \dfrac{1}{{{{\left( {x - 1} \right)}^2}}}\\
\to \dfrac{{{x^2} - 2x + 1 - x - 1}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} < 0\\
\to \dfrac{{{x^2} - 3x}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} < 0\\
\to \dfrac{{x\left( {x - 3} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} < 0
\end{array}\)
BXD:
x -∞ -1 0 1(kép) 3 +∞
f(x) - // + 0 - // - 0 +
\(KL:x \in \left( { - \infty ; - 1} \right) \cup \left( {0;1} \right) \cup \left( {1;3} \right)\)
\(\begin{array}{l}
C16:\\
DK:x \ne \pm 3\\
\dfrac{{x + 4}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \dfrac{2}{{x + 3}} < \dfrac{{4x}}{{x\left( {3 - x} \right)}}\\
\to \dfrac{{x + 4}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \dfrac{2}{{x + 3}} < \dfrac{{ - 4}}{{x - 3}}\\
\to \dfrac{{x + 4 - 2x + 6 + 4x + 12}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} < 0\\
\to \dfrac{{3x + 22}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} < 0
\end{array}\)
BXD:
x -∞ -22/3 -3 3 +∞
f(x) - 0 + // - // +
\(\begin{array}{l}
\to x \in \left( { - \infty ; - \dfrac{{22}}{3}} \right) \cup \left( { - 3;3} \right)\\
\to {x_{\max }} = 2
\end{array}\)
