Đáp án:
b) \(x = \dfrac{7}{9}\)
Giải thích các bước giải:
\(\begin{array}{l}
C4:\\
a)Ycbt \to \left\{ \begin{array}{l}
m \ne - 1\\
\Delta ' > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne - 1\\
{m^2} - 2m + 1 - \left( {m + 1} \right)\left( {m - 2} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne - 1\\
{m^2} - 2m + 1 - {m^2} + m + 2 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne - 1\\
- m + 3 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne - 1\\
3 > m
\end{array} \right.\\
b)Thay:x = 3\\
Pt \to \left( {m + 1} \right).9 - 2\left( {m - 1} \right).3 + m - 2 = 0\\
\to 9m + 9 - 6m + 6 + m - 2 = 0\\
\to 4m = - 13\\
\to m = - \dfrac{{13}}{4}\\
Thay:m = - \dfrac{{13}}{4}\\
Pt \to - \dfrac{9}{4}{x^2} + \dfrac{{17}}{2}x - \dfrac{{21}}{4} = 0\\
\to - 9{x^2} + 34x - 21 = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = \dfrac{7}{9}
\end{array} \right.
\end{array}\)
