xét dấu mình bài này với : 1/x+2 < 1/( x -2 )^ 08/07/2021 Bởi Daisy xét dấu mình bài này với : 1/x+2 < 1/( x -2 )^
Đáp án: \(x \in \left( { – \infty ; – 2} \right) \cup \left( {\dfrac{{5 – \sqrt {17} }}{2};2} \right) \cup \left( {2;\dfrac{{5 + \sqrt {17} }}{2}} \right)\) Giải thích các bước giải: \(\begin{array}{l}DK:x \ne \pm 2\\\dfrac{1}{{x + 2}} < \dfrac{1}{{{{\left( {x – 2} \right)}^2}}}\\ \to \dfrac{{{x^2} – 4x + 4 – x – 2}}{{\left( {x + 2} \right){{\left( {x – 2} \right)}^2}}} < 0\\ \to \dfrac{{{x^2} – 5x + 2}}{{\left( {x + 2} \right){{\left( {x – 2} \right)}^2}}} < 0\\Xét:\dfrac{{{x^2} – 5x + 2}}{{\left( {x + 2} \right){{\left( {x – 2} \right)}^2}}} = 0\\ \to {x^2} – 5x + 2 = 0\\ \to \left[ \begin{array}{l}x = \dfrac{{5 + \sqrt {17} }}{2}\\x = \dfrac{{5 – \sqrt {17} }}{2}\end{array} \right.\end{array}\) BXD: x -∞ -2 \(\dfrac{{5 – \sqrt {17} }}{2}\) 2(kép) \(\dfrac{{5 – \sqrt {17} }}{2}\) +∞ f(x) – // + 0 – // – 0 + \(KL:x \in \left( { – \infty ; – 2} \right) \cup \left( {\dfrac{{5 – \sqrt {17} }}{2};2} \right) \cup \left( {2;\dfrac{{5 + \sqrt {17} }}{2}} \right)\) Bình luận
Đáp án:
\(x \in \left( { – \infty ; – 2} \right) \cup \left( {\dfrac{{5 – \sqrt {17} }}{2};2} \right) \cup \left( {2;\dfrac{{5 + \sqrt {17} }}{2}} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \pm 2\\
\dfrac{1}{{x + 2}} < \dfrac{1}{{{{\left( {x – 2} \right)}^2}}}\\
\to \dfrac{{{x^2} – 4x + 4 – x – 2}}{{\left( {x + 2} \right){{\left( {x – 2} \right)}^2}}} < 0\\
\to \dfrac{{{x^2} – 5x + 2}}{{\left( {x + 2} \right){{\left( {x – 2} \right)}^2}}} < 0\\
Xét:\dfrac{{{x^2} – 5x + 2}}{{\left( {x + 2} \right){{\left( {x – 2} \right)}^2}}} = 0\\
\to {x^2} – 5x + 2 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{5 + \sqrt {17} }}{2}\\
x = \dfrac{{5 – \sqrt {17} }}{2}
\end{array} \right.
\end{array}\)
BXD:
x -∞ -2 \(\dfrac{{5 – \sqrt {17} }}{2}\) 2(kép) \(\dfrac{{5 – \sqrt {17} }}{2}\) +∞
f(x) – // + 0 – // – 0 +
\(KL:x \in \left( { – \infty ; – 2} \right) \cup \left( {\dfrac{{5 – \sqrt {17} }}{2};2} \right) \cup \left( {2;\dfrac{{5 + \sqrt {17} }}{2}} \right)\)