Giải bất phương trình: x+1/x²-4x-5>3/2x+1 29/09/2021 Bởi Audrey Giải bất phương trình: x+1/x²-4x-5>3/2x+1
Đáp án: \(x \in \left( { – \infty ; – 1} \right) \cup \left( { – 1; – \dfrac{1}{2}} \right) \cup \left( {5;16} \right)\) Giải thích các bước giải: \(\begin{array}{l}DK:x \ne \left\{ { – 1; – \dfrac{1}{2};5} \right\}\\\dfrac{{x + 1}}{{{x^2} – 4x – 5}} > \dfrac{3}{{2x + 1}}\\ \to \dfrac{{\left( {x + 1} \right)\left( {2x + 1} \right) – 3\left( {{x^2} – 4x – 5} \right)}}{{\left( {x + 1} \right)\left( {x – 5} \right)\left( {2x + 1} \right)}} > 0\\ \to \dfrac{{2{x^2} + 3x + 1 – 3{x^2} + 12x + 15}}{{\left( {x + 1} \right)\left( {x – 5} \right)\left( {2x + 1} \right)}} > 0\\ \to \dfrac{{ – {x^2} + 15x + 16}}{{\left( {x + 1} \right)\left( {x – 5} \right)\left( {2x + 1} \right)}} > 0\\ \to \dfrac{{\left( {16 – x} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x – 5} \right)\left( {2x + 1} \right)}} > 0\end{array}\) BXD: x -∞ -1(kép) -1/2 5 16 +∞ f(x) + // + // – // + 0 – \(KL:x \in \left( { – \infty ; – 1} \right) \cup \left( { – 1; – \dfrac{1}{2}} \right) \cup \left( {5;16} \right)\) Bình luận
Đáp án:
\(x \in \left( { – \infty ; – 1} \right) \cup \left( { – 1; – \dfrac{1}{2}} \right) \cup \left( {5;16} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \left\{ { – 1; – \dfrac{1}{2};5} \right\}\\
\dfrac{{x + 1}}{{{x^2} – 4x – 5}} > \dfrac{3}{{2x + 1}}\\
\to \dfrac{{\left( {x + 1} \right)\left( {2x + 1} \right) – 3\left( {{x^2} – 4x – 5} \right)}}{{\left( {x + 1} \right)\left( {x – 5} \right)\left( {2x + 1} \right)}} > 0\\
\to \dfrac{{2{x^2} + 3x + 1 – 3{x^2} + 12x + 15}}{{\left( {x + 1} \right)\left( {x – 5} \right)\left( {2x + 1} \right)}} > 0\\
\to \dfrac{{ – {x^2} + 15x + 16}}{{\left( {x + 1} \right)\left( {x – 5} \right)\left( {2x + 1} \right)}} > 0\\
\to \dfrac{{\left( {16 – x} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x – 5} \right)\left( {2x + 1} \right)}} > 0
\end{array}\)
BXD:
x -∞ -1(kép) -1/2 5 16 +∞
f(x) + // + // – // + 0 –
\(KL:x \in \left( { – \infty ; – 1} \right) \cup \left( { – 1; – \dfrac{1}{2}} \right) \cup \left( {5;16} \right)\)