Đáp án:
\(\begin{array}{l}
a)\,\,m = 19;\,\,{x_2} = \frac{5}{8}\\
b)\,\,m > - \frac{7}{6},\,\,m \ne 3\\
c)\,\,m = - \frac{3}{4}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left( {m - 3} \right){x^2} - 2\left( {m + 2} \right)x + m + 1 = 0\\
a)\,\,{x_1} = 2\,\,\,la\,\,nghiem\,\,cua\,\,pt\\
\Rightarrow 4\left( {m - 3} \right) - 2\left( {m + 2} \right).2 + m + 1 = 0\\
\Leftrightarrow 4m - 12 - 4m - 8 + m + 1 = 0\\
\Leftrightarrow m - 19 = 0 \Leftrightarrow m = 19\\
Khi\,\,m = 19,\,\,pt:\,\,16{x^2} - 42x + 20 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \frac{5}{8}
\end{array} \right.\\
b)\,\,\Delta ' = {\left( {m + 2} \right)^2} - \left( {m - 3} \right)\left( {m + 1} \right)\\
\,\,\,\,\,\,\,\,\,\,\,\, = {m^2} + 4m + 4 - {m^2} + 2m + 3\\
\,\,\,\,\,\,\,\,\,\,\,\, = 6m + 7\\
Pt\,\,co\,\,2\,\,nghiem\,\,pb\,\, \Leftrightarrow \left\{ \begin{array}{l}
\Delta ' > 0\\
a \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
6m + 7 > 0\\
m \ne 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m > - \frac{7}{6}\\
m \ne 3
\end{array} \right.\\
c)\,\,Khi\,\,m > - \frac{7}{6},\,\,m \ne 3,\,\,ap\,\,dung\,\,dinh\,\,li\,\,Vi - et:\\
\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{2\left( {m + 2} \right)}}{{m - 3}}\\
{x_1}{x_2} = \frac{{m + 1}}{{m - 3}}
\end{array} \right.\\
Theo\,\,gt:\,\,\frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} = 10\\
\Leftrightarrow \frac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} = 10 \Leftrightarrow {x_1} + {x_2} = 10{x_1}{x_2}\\
\Leftrightarrow \frac{{2\left( {m + 2} \right)}}{{m - 3}} = 10\frac{{m + 1}}{{m - 3}}\\
\Leftrightarrow 2m + 4 = 10m + 10\\
\Leftrightarrow 8m = - 6 \Leftrightarrow m = - \frac{3}{4}\,\,\left( {tm} \right)
\end{array}\)
