Cho f(x)=(3m +1)x-(3m +1)x + m +4 (m là tham số)
1) Tìm m để f(x) <0 với mọi x thuộc R.
2) Tìm m để f(x)>0 với mọi x thuộc R.
3) Tìm m để phương trình f(x) = 0 có hai nghiệm dương phân biệt.
4) Tìm m để phương trình f(x) = 0 có hai nghiệm phân biệt x :x, thỏa mãn
Đáp án:
c. \(m \in \left( { – 15; – 4} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left\{ \begin{array}{l}
3m + 1 < 0\\
9{m^2} + 6m + 1 – 4.\left( {3m + 1} \right)\left( {m + 4} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < – \dfrac{1}{3}\\
9{m^2} + 6m + 1 – \left( {12m + 4} \right)\left( {m + 4} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < – \dfrac{1}{3}\\
9{m^2} + 6m + 1 – 12{m^2} – 52m – 16 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < – \dfrac{1}{3}\\
– 3{m^2} – 46m – 15 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < – \dfrac{1}{3}\\
– \left( {3m + 1} \right)\left( {m + 15} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < – \dfrac{1}{3}\\
\left( {3m + 1} \right)\left( {m + 15} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < – \dfrac{1}{3}\\
m \in \left( { – \infty ; – 15} \right) \cup \left( { – \dfrac{1}{3}; + \infty } \right)
\end{array} \right.\\
\to m \in \left( { – \infty ; – 15} \right)\\
b.\left\{ \begin{array}{l}
3m + 1 > 0\\
9{m^2} + 6m + 1 – 4.\left( {3m + 1} \right)\left( {m + 4} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > – \dfrac{1}{3}\\
9{m^2} + 6m + 1 – \left( {12m + 4} \right)\left( {m + 4} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > – \dfrac{1}{3}\\
9{m^2} + 6m + 1 – 12{m^2} – 52m – 16 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > – \dfrac{1}{3}\\
– 3{m^2} – 46m – 15 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > – \dfrac{1}{3}\\
– \left( {3m + 1} \right)\left( {m + 15} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > – \dfrac{1}{3}\\
\left( {3m + 1} \right)\left( {m + 15} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > – \dfrac{1}{3}\\
m \in \left( { – \infty ; – 15} \right) \cup \left( { – \dfrac{1}{3}; + \infty } \right)
\end{array} \right.\\
\to m \in \left( { – \dfrac{1}{3}; + \infty } \right)\\
c.Ycbt \to \left\{ \begin{array}{l}
9{m^2} + 6m + 1 – 4.\left( {3m + 1} \right)\left( {m + 4} \right) > 0\\
\dfrac{{m + 4}}{{3m + 1}} > 0\\
\dfrac{{3m + 1}}{{3m + 1}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
– \left( {3m + 1} \right)\left( {m + 15} \right) > 0\\
m \in \left( { – \infty ; – 4} \right) \cup \left( { – \dfrac{1}{3}; + \infty } \right)\\
1 > 0\left( {ld} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {3m + 1} \right)\left( {m + 15} \right) < 0\\
m \in \left( { – \infty ; – 4} \right) \cup \left( { – \dfrac{1}{3}; + \infty } \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \in \left( { – 15; – \dfrac{1}{3}} \right)\\
m \in \left( { – \infty ; – 4} \right) \cup \left( { – \dfrac{1}{3}; + \infty } \right)
\end{array} \right.\\
\to m \in \left( { – 15; – 4} \right)
\end{array}\)