Đáp án:
2.b. \(m > \frac{7}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
1.Đặt:{x^2} + x = t\\
Pt \to 3{t^2} - 2t - 1 = 0\\
\to 3{t^2} - 3t + t - 1 = 0\\
\to 3t\left( {t - 1} \right) + t - 1 = 0\\
\to \left[ \begin{array}{l}
t - 1 = 0\\
3t + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
t = 1\\
t = - \frac{1}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} + x = 1\\
{x^2} + x = - \frac{1}{3}\left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \frac{{ - 1 + \sqrt 5 }}{2}\\
x = \frac{{ - 1 - \sqrt 5 }}{2}
\end{array} \right.\\
2.a. Δ= {m^2} - 4\left( {m - 2} \right)\\
= {m^2} - 4m + 8\\
= {\left( {m - 2} \right)^2} + 4 > 0\left( {ld} \right)\forall m \in R
\end{array}\)
⇒ Phương trình luôn có 2 nghiệm phân biệt với mọi m
\(\begin{array}{l}
b.Do:\left\{ \begin{array}{l}
{x_1} > \frac{1}{2}\\
{x_2} > \frac{1}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x_1} - \frac{1}{2} > 0\\
{x_2} - \frac{1}{2} > 0
\end{array} \right.\\
\to \left( {{x_1} - \frac{1}{2}} \right)\left( {{x_2} - \frac{1}{2}} \right) > 0\\
\to {x_1}{x_2} - \frac{1}{2}\left( {{x_1} + {x_2}} \right) + \frac{1}{4} > 0\\
\to m - 2 - \frac{1}{2}.m + \frac{1}{4} > 0\\
\to \frac{1}{2}m > \frac{7}{4}\\
\to m > \frac{7}{2}
\end{array}\)
