Đáp án:
c) \(\left[ \begin{array}{l}
m = 2\\
m = \dfrac{{31}}{{28}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:m = 4\\
Pt \to 2{x^2} + 9x + 4 = 4\\
\to 2{x^2} + 9x = 0\\
\to x\left( {2x + 9} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - \dfrac{9}{2}
\end{array} \right.\\
b)\left( {m - 2} \right){x^2} + \left( {2m + 1} \right)x + m - 4 = 0
\end{array}\)
Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
m \ne 2\\
4{m^2} + 4m + 1 - 4\left( {m - 2} \right)\left( {m - 4} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 2\\
4{m^2} + 4m + 1 - 4\left( {{m^2} - 6m + 8} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 2\\
4{m^2} + 4m + 1 - 4{m^2} + 24m - 32 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 2\\
28m - 31 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 2\\
m > \dfrac{{31}}{{28}}
\end{array} \right.\\
Có:{x_1}{x_2} = 3\\
\to \dfrac{{m - 4}}{{m - 2}} = 3\\
\to m - 4 = 2m - 6\\
\to m = 2\left( l \right)\\
\to m \in \emptyset \\
c)TH1:m - 2 = 0 \to m = 2\\
Thay:m = 2\\
Pt \to 5x + 2 = 4\\
\to x = \dfrac{2}{5}\left( {TMDK} \right)\\
\to m = 2\left( {TM} \right)\\
TH2:m \ne 2\\
Pt:\left( {m - 2} \right){x^2} + \left( {2m + 1} \right)x + m - 4 = 0\\
Ycbt \Leftrightarrow \Delta = 0\\
\to 4{m^2} + 4m + 1 - 4\left( {m - 2} \right)\left( {m - 4} \right) = 0\\
\to 4{m^2} + 4m + 1 - 4\left( {{m^2} - 6m + 8} \right) = 0\\
\to 4{m^2} + 4m + 1 - 4{m^2} + 24m - 32 = 0\\
\to 28m - 31 = 0\\
\to m = \dfrac{{31}}{{28}}
\end{array}\)
\(KL:\left[ \begin{array}{l}
m = 2\\
m = \dfrac{{31}}{{28}}
\end{array} \right.\)
