Đáp án:
2) a) m=2
b) m<7/8
Giải thích các bước giải:
\(\begin{array}{l}
1b)\,TH1:\,m = 0 \Rightarrow - 4x + 3 = 0 \Leftrightarrow x = \frac{3}{4}\\
TH2:\,m \ne 0\\
\Delta ' = {\left( {m - 2} \right)^2} - m\left( {m + 3} \right) = - 7m + 1\\
\Delta ' > 0 \Leftrightarrow m < \frac{1}{7} \Rightarrow pt\,co\,hai\,nghiem\,phan\,biet\\
{x_{1,2}} = 2 - m \pm \sqrt { - 7m + 1} \\
\Delta ' = 0 \Leftrightarrow m = \frac{1}{7} \Rightarrow pt\,co\,nghiem\,kep\,{x_1} = {x_2} = \frac{{2 - n}}{m}\\
\Delta ' < 0 \Rightarrow m > \frac{1}{7} \Rightarrow pt\,VN\\
2)\,a)\,\left( {{m^2} - 4} \right)x = - 2\left( {m + 2} \right)\\
PT\,VN\, \Leftrightarrow \left\{ \begin{array}{l}
{m^2} - 4 = 0\\
- 2\left( {m + 2} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m = \pm 2\\
m \ne - 2
\end{array} \right. \Leftrightarrow m = 2\\
b)\,TH1:\,m = 1 \Leftrightarrow - x - 1 = 0 \Leftrightarrow x = - 1\left( L \right)\\
TH2:\,m \ne 1\\
\Delta = {\left( {2m - 1} \right)^2} - 4\left( {m - 1} \right)\left( {m - 2} \right)\\
= 4{m^2} - 4m + 1 - 4\left( {{m^2} - 3m + 2} \right)\\
= 8m - 7 < 0\\
\Leftrightarrow m < \frac{7}{8}\,thi\,pt\,VN
\end{array}\)
